3.8 Solving trigonometric equations · IB Maths AA

Key terms introduced here

Each is in the Maths AA glossary, with every other term in the subject.

Trigonometric equation: An equation involving circular functions of an unknown angle, such as $2\sin x = 1$ . It usually has infinitely many solutions, so it is solved on a stated interval.

Notes on 3.8 Solving trigonometric equations

26 min readLast reviewed 14 Oct 2026

A trigonometric equation asks for the angles at which a circular function takes a given value: the times when the tide is 2 m deep, the angles at which a projectile reaches a given range, the moments when a wheel's capsule is level with a platform. Because circular functions repeat, such an equation usually has infinitely many solutions, and a question always says which interval of angles to search.

Solving one well uses three ideas together: the calculator (or an exact value) gives one solution, the symmetry of the unit circle gives a second, and the period of the function gives all the others. More complicated equations are reduced to these simple ones by algebra.

What it means to solve a trigonometric equation

Every time the graph of $y = \sin x$ crosses the line $y = 0.5$ , there is a solution of $\sin x = 0.5$ . The sine curve keeps rising and falling forever, so it crosses the line twice in every cycle of length $2\pi$ , and there are infinitely many solutions.

y = sin x

The line y = 0.5 meets the sine curve twice in every cycle. Each pair of solutions is 2π further on than the previous pair.

The solutions in this graph are

x = \frac{\pi}{6} + 2k\pi \quad \text{or} \quad x = \frac{5\pi}{6} + 2k\pi, \qquad k \in \mathbb{Z}

A question restricts the answer to an interval, such as $0 \le x \le 2\pi$ or $-180^\circ < x \le 180^\circ$ , and the task is to list exactly the solutions inside it: no more and no fewer.

The value given by a calculator

The inverse functions on a calculator, written $\sin^{-1}$ , $\cos^{-1}$ and $\tan^{-1}$ , return a single angle:

Function	Output range (radians)	Output range (degrees)
$\sin^{-1} k$	$-\frac{\pi}{2} \le x \le \frac{\pi}{2}$	$-90^\circ \le x \le 90^\circ$
$\cos^{-1} k$	$0 \le x \le \pi$	$0^\circ \le x \le 180^\circ$
$\tan^{-1} k$	$-\frac{\pi}{2} < x < \frac{\pi}{2}$	$-90^\circ < x < 90^\circ$

This single value is one solution out of infinitely many. It may not even lie in the interval the question asks about, so it is a starting point, not the answer.

Solving sin x = k, cos x = k and tan x = k

The second solution in each cycle comes from the symmetry of the unit circle. The point at angle $x$ on the unit circle is $(\cos x, \sin x)$ , so an equation in $\sin x$ fixes the height of the point, an equation in $\cos x$ fixes its horizontal position, and an equation in $\tan x$ fixes the gradient of the radius.

Three unit circles. sin x = k: the line y = k meets the circle at angles α and π − α. cos x = k: the line x = k meets it at α and 2π − α. tan x = k: a line through O meets it at α and π + α. Each pair makes equal angles α with the x-axis. — Each equation is met at two points of the circle that make the same angle α with the x-axis: mirror images in the y-axis for sine, in the x-axis for cosine, and opposite ends of a diameter for tangent.

$\sin x = k$ : the horizontal line $y = k$ meets the circle at two points that are reflections of each other in the $y$ -axis. If one is at angle $\alpha$ , the other is at $\pi - \alpha$ .
$\cos x = k$ : the vertical line $x = k$ meets the circle at two points that are reflections in the $x$ -axis, at angles $\alpha$ and $-\alpha$ . Within $0 \le x \le 2\pi$ , the second is written $2\pi - \alpha$ .
$\tan x = k$ : the line through $O$ with gradient $k$ meets the circle at the two ends of a diameter, at angles $\alpha$ and $\pi + \alpha$ .

All other solutions come from adding or subtracting whole periods: $2\pi$ for sine and cosine, $\pi$ for tangent. In degrees, the same rules read $180^\circ - \alpha$ , $360^\circ - \alpha$ and $180^\circ + \alpha$ , with periods $360^\circ$ and $180^\circ$ .

Equation	First solution $\alpha$	Second solution	Add multiples of
$\sin x = k$ , $-1 \le k \le 1$	$\sin^{-1} k$	$\pi - \alpha$	$2\pi$
$\cos x = k$ , $-1 \le k \le 1$	$\cos^{-1} k$	$2\pi - \alpha$ (or $-\alpha$ )	$2\pi$
$\tan x = k$ , any $k$	$\tan^{-1} k$	$\pi + \alpha$	$\pi$ (the second solution is already one period on)

A method for every interval

Rearrange the equation to the form $\sin x = k$ , $\cos x = k$ or $\tan x = k$ .
Find one solution $\alpha$ , exactly (from the unit circle and the related angle, 3.5) or with the calculator.
Use the symmetry rule to find the second solution in the same cycle.
Add or subtract the period to both solutions until every value in the interval has been found, and discard any outside it.
Check the count: a sine or cosine equation with $-1 < k < 1$ has two solutions in each full cycle.

A sketch of the graph and the horizontal line $y = k$ over the interval shows how many solutions to expect before any calculation.

The method works in the same way when the value is negative, but the calculator value often lies outside the interval and has to be moved into it first.

Solve $\tan x = -\sqrt{3}$ for $0 \le x \le 2\pi$ .

Show the solution

The related angle is $\frac{\pi}{3}$ , since $\tan\frac{\pi}{3} = \sqrt{3}$ . Tangent is negative in the second and fourth quadrants.

One solution is $\alpha = -\frac{\pi}{3}$ (the calculator value), which lies outside the interval. Adding the period $\pi$ repeatedly:

-\frac{\pi}{3} + \pi = \frac{2\pi}{3} \qquad -\frac{\pi}{3} + 2\pi = \frac{5\pi}{3}

The next value, $\frac{8\pi}{3}$ , is greater than $2\pi$ .

x = \frac{2\pi}{3} \quad \text{or} \quad x = \frac{5\pi}{3}

Check: $\frac{2\pi}{3}$ is in the second quadrant and $\frac{5\pi}{3}$ is in the fourth, where tangent is negative.

For a cosine equation, the second solution is the negative of the first. This form is the most convenient when the interval is centred on zero, such as $-\pi \le x \le \pi$ .

When the value of $k$ is not one of the exact values, the calculator gives $\alpha$ as a decimal. Keep at least five significant figures in working and round only at the end.

Solve the following equations, giving answers correct to 3 significant figures.

(a) $3\cos x = 2$ for $0 \le x \le 2\pi$

(b) $\sin x = -0.4$ for $0 \le x \le 2\pi$

Show the solution

(a) $\cos x = \frac{2}{3}$ , so $\alpha = \cos^{-1}\frac{2}{3} = 0.84106\ldots$ The second solution is $2\pi - 0.84106\ldots = 5.44211\ldots$

x = 0.841 \quad \text{or} \quad x = 5.44

(b) $\alpha = \sin^{-1}(-0.4) = -0.41151\ldots$ , which is outside the interval. The second solution in the same cycle is

\pi - (-0.41151\ldots) = 3.55311\ldots

and adding $2\pi$ to the first gives $-0.41151\ldots + 2\pi = 5.87166\ldots$

x = 3.55 \quad \text{or} \quad x = 5.87

Both lie in the third and fourth quadrants, where sine is negative.

When the interval is in degrees, the same rules apply with $180^\circ$ and $360^\circ$ . For $\sin\theta = 0.3$ , $0^\circ \le \theta \le 360^\circ$ , the calculator in degree mode gives $\theta = 17.5^\circ$ , and the second solution is $180^\circ - 17.5^\circ = 162.5^\circ$ . Angles in degrees are given to 1 decimal place.

Values with only one solution per cycle

When $k = 1$ , $k = -1$ or $k = 0$ , the two symmetric points on the circle coincide or sit at the ends of the axes, and the general rules can produce repeated or misleading values. These cases are best read straight from the graph:

$\sin x = 1$ only at $x = \frac{\pi}{2} + 2k\pi$ ; $\sin x = -1$ only at $x = \frac{3\pi}{2} + 2k\pi$ ;
$\cos x = 1$ only at $x = 2k\pi$ ; $\cos x = -1$ only at $x = \pi + 2k\pi$ ;
$\sin x = 0$ at $x = k\pi$ ; $\cos x = 0$ at $x = \frac{\pi}{2} + k\pi$ ; $\tan x = 0$ at $x = k\pi$ .

For example, $\cos x = 1$ for $0 \le x \le 2\pi$ has the two solutions $x = 0$ and $x = 2\pi$ , both end points of the interval, while $\sin x = 1$ on the same interval has only $x = \frac{\pi}{2}$ .

Solving trigonometric equations graphically

Some trigonometric equations mix different functions or angles in a way that has no algebraic method at this level, such as $3\sin 2x + 1 = 2\cos x$ . A GDC solves them graphically: graph each side as a separate function and find the $x$ -coordinates of the points of intersection. Alternatively, graph the difference of the two sides and find its zeros.

Equations with a transformed angle

An equation such as $\sin 2x = 0.5$ involves the function of $2x$ , not of $x$ . The graph of $y = \sin 2x$ has period $\pi$ , so it completes two cycles for every one cycle of $y = \sin x$ , and the equation has twice as many solutions in $0 \le x \le 2\pi$ .

The method is to treat the whole angle as a new unknown, solve for that, and convert back:

Let $u$ be the angle inside the function (for example $u = 2x$ , or $u = x - \frac{\pi}{3}$ ).
Transform the interval for $x$ into an interval for $u$ , by applying to its end points the same operations that turn $x$ into $u$ .
Solve the simple equation for $u$ on the new interval, finding every solution.
Convert each value of $u$ back to $x$ .

Solve $\sin 2x = 0.5$ for $0 \le x \le 2\pi$ .

Show the solution

Let $u = 2x$ . Since $0 \le x \le 2\pi$ , multiplying by $2$ gives $0 \le u \le 4\pi$ : two full cycles.

Solve $\sin u = 0.5$ for $0 \le u \le 4\pi$ . The first cycle gives $u = \frac{\pi}{6}$ and $u = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$ . Adding $2\pi$ gives the second cycle:

u = \frac{\pi}{6}, \ \frac{5\pi}{6}, \ \frac{13\pi}{6}, \ \frac{17\pi}{6}

Divide by $2$ to return to $x$ :

x = \frac{\pi}{12}, \ \frac{5\pi}{12}, \ \frac{13\pi}{12}, \ \frac{17\pi}{12}

y = sin 2x

y = sin 2x completes two cycles between 0 and 2π, so the line y = 0.5 crosses it four times.

The same method handles a translation inside the function. The interval is shifted rather than stretched, and some solutions of the simple equation may fall outside the new interval.

Solve $2\cos\left(x - \frac{\pi}{3}\right) = -1$ for $0 \le x \le 2\pi$ .

Show the solution

$\cos\left(x - \frac{\pi}{3}\right) = -\frac{1}{2}$ . Let $u = x - \frac{\pi}{3}$ . Subtracting $\frac{\pi}{3}$ from the end points gives

-\frac{\pi}{3} \le u \le \frac{5\pi}{3}

Solve $\cos u = -\frac{1}{2}$ : the related angle is $\frac{\pi}{3}$ and cosine is negative in the second and third quadrants, so $u = \frac{2\pi}{3}$ or $u = \frac{4\pi}{3}$ . Both lie in the interval; the next values, $\frac{2\pi}{3} - 2\pi$ and $\frac{4\pi}{3} - 2\pi = -\frac{2\pi}{3}$ , are below $-\frac{\pi}{3}$ .

Add $\frac{\pi}{3}$ to return to $x$ :

x = \frac{2\pi}{3} + \frac{\pi}{3} = \pi \quad \text{or} \quad x = \frac{4\pi}{3} + \frac{\pi}{3} = \frac{5\pi}{3}

The interval for $u$ can also be larger than one cycle and start below zero, as in the next two examples.

Solve the following equations.

(a) $\tan 3x = 1$ for $0 \le x \le \pi$

(b) $2\cos 2x + 1 = 0$ for $-\pi \le x \le \pi$

Show the solution

(a) Let $u = 3x$ , so $0 \le u \le 3\pi$ . The equation $\tan u = 1$ has $\alpha = \frac{\pi}{4}$ and period $\pi$ :

u = \frac{\pi}{4}, \ \frac{5\pi}{4}, \ \frac{9\pi}{4}

( $\frac{13\pi}{4} > 3\pi$ .) Dividing by $3$ :

x = \frac{\pi}{12}, \ \frac{5\pi}{12}, \ \frac{3\pi}{4}

(b) $\cos 2x = -\frac{1}{2}$ . Let $u = 2x$ , so $-2\pi \le u \le 2\pi$ . In $0 \le u \le 2\pi$ , $\cos u = -\frac{1}{2}$ gives $u = \frac{2\pi}{3}$ and $\frac{4\pi}{3}$ . Subtracting $2\pi$ gives the solutions in $-2\pi \le u \le 0$ : $u = -\frac{4\pi}{3}$ and $-\frac{2\pi}{3}$ . Dividing by $2$ :

x = -\frac{2\pi}{3}, \ -\frac{\pi}{3}, \ \frac{\pi}{3}, \ \frac{2\pi}{3}

The four answers are symmetrical about $0$ , as expected for an equation in $\cos 2x$ .

In degrees, the working is identical. For $\sin(2\theta - 30^\circ) = 0.5$ with $0^\circ \le \theta \le 360^\circ$ , the new angle $u = 2\theta - 30^\circ$ runs from $-30^\circ$ to $690^\circ$ . The solutions $u = 30^\circ, 150^\circ, 390^\circ, 510^\circ$ give $\theta = 30^\circ, 90^\circ, 210^\circ, 270^\circ$ .

Equations from sinusoidal models

A sinusoidal model from 3.7 leads to exactly this kind of equation whenever a question asks when the modelled quantity reaches a given value. In a simplified model of the London Eye, the height of a capsule $t$ minutes after it leaves the lowest point is $h(t) = 60 - 60\cos\left(\frac{\pi t}{15}\right)$ metres, with one revolution every $30$ minutes.

Find, for one revolution of the wheel, the length of time for which the capsule is more than $90$ m above the ground.

Show the solution

First solve $h(t) = 90$ for $0 \le t \le 30$ :

60 - 60\cos\left(\frac{\pi t}{15}\right) = 90 \quad \Rightarrow \quad \cos\left(\frac{\pi t}{15}\right) = -\frac{1}{2}

Let $u = \frac{\pi t}{15}$ ; for $0 \le t \le 30$ , $0 \le u \le 2\pi$ . Then $u = \frac{2\pi}{3}$ or $u = \frac{4\pi}{3}$ , so

t = \frac{15}{\pi} \times \frac{2\pi}{3} = 10 \quad \text{or} \quad t = \frac{15}{\pi} \times \frac{4\pi}{3} = 20

The capsule rises through $90$ m after $10$ minutes and falls back through it after $20$ minutes; the highest point ( $t = 15$ ) lies between. It is above $90$ m for $20 - 10 = 10$ minutes, one third of each revolution.

Equations leading to quadratic equations

Some trigonometric equations are quadratic equations in disguise. In $2\sin^2 x - \sin x - 1 = 0$ , the unknown $\sin x$ plays the part of $y$ in $2y^2 - y - 1 = 0$ . Such an equation in quadratic form is solved in two stages: first solve the quadratic for $\sin x$ (2.7), then solve each resulting simple equation for $x$ .

Remember the notation: $\sin^2 x$ means $(\sin x)^2$ .

Solve $2\sin^2 x - \sin x - 1 = 0$ for $0 \le x \le 2\pi$ .

Show the solution

Factorise the quadratic in $\sin x$ (it may help to write $s = \sin x$ , giving $2s^2 - s - 1 = 0$ ):

(2\sin x + 1)(\sin x - 1) = 0

So $\sin x = -\frac{1}{2}$ or $\sin x = 1$ .

$\sin x = 1$ gives $x = \frac{\pi}{2}$ only.

$\sin x = -\frac{1}{2}$ : the related angle is $\frac{\pi}{6}$ , and sine is negative in the third and fourth quadrants, so $x = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$ or $x = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$ .

x = \frac{\pi}{2}, \ \frac{7\pi}{6}, \ \frac{11\pi}{6}

Rejecting impossible values

One root of the quadratic may be a value that sine or cosine can never take. That root gives no solutions and is rejected with a reason.

When the quadratic does not factorise, use the quadratic formula, then a calculator for the angles.

Solve $\tan^2 x - 3\tan x + 1 = 0$ for $0 \le x \le \pi$ , giving your answers correct to 3 significant figures.

Show the solution

By the quadratic formula with $\tan x$ as the unknown,

\tan x = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2}

so $\tan x = 2.6180\ldots$ or $\tan x = 0.38196\ldots$

Both values are positive, so the solutions lie in the first quadrant (and in the third, which is outside $0 \le x \le \pi$ ). Adding the period $\pi$ to either first-quadrant value takes it beyond $\pi$ .

x = \tan^{-1}(2.6180\ldots) = 1.21 \quad \text{or} \quad x = \tan^{-1}(0.38196\ldots) = 0.365

Common factors: never divide by a function that could be zero

In an equation such as $2\sin x\cos x = \sin x$ , dividing both sides by $\sin x$ looks tempting. It loses every solution where $\sin x = 0$ . Move every term to one side and factorise instead.

Using identities to get one function

An equation that contains two different circular functions, or both $x$ and $2x$ , can often be rewritten in terms of a single function of $x$ with an identity from 3.6. The aim is either a quadratic in one function or a product of simple factors.

The equation contains	Identity to use	Result
$\cos^2 x$ and $\sin x$	$\cos^2 x = 1 - \sin^2 x$	Quadratic in $\sin x$
$\sin^2 x$ and $\cos x$	$\sin^2 x = 1 - \cos^2 x$	Quadratic in $\cos x$
$\sin 2x$	$\sin 2x = 2\sin x\cos x$	Usually a common factor
$\cos 2x$ and $\sin x$	$\cos 2x = 1 - 2\sin^2 x$	Quadratic in $\sin x$
$\cos 2x$ and $\cos x$	$\cos 2x = 2\cos^2 x - 1$	Quadratic in $\cos x$
$\sin x$ and $\cos x$ , both to the first power and no constant	divide by $\cos x$ (when $\cos x \neq 0$ )	$\tan x = k$

The Pythagorean identity

The Pythagorean identity $\cos^2 x + \sin^2 x = 1$ swaps a squared function for the other one.

The same substitution works the other way round, turning $\sin^2 x$ into an expression in $\cos x$ .

The double angle identities

The double angle identities replace a function of $2x$ by functions of $x$ . For $\cos 2x$ , choose the one of its three forms that leaves only one function in the equation.

Solve $\sin 2x = \cos x$ for $0 \le x \le 2\pi$ .

Show the solution

Use $\sin 2x = 2\sin x\cos x$ and collect everything on one side:

2\sin x\cos x - \cos x = 0 \quad \Rightarrow \quad \cos x(2\sin x - 1) = 0

So $\cos x = 0$ or $\sin x = \frac{1}{2}$ .

$\cos x = 0$ : $x = \frac{\pi}{2}$ or $x = \frac{3\pi}{2}$ .

$\sin x = \frac{1}{2}$ : $x = \frac{\pi}{6}$ or $x = \frac{5\pi}{6}$ .

x = \frac{\pi}{6}, \ \frac{\pi}{2}, \ \frac{5\pi}{6}, \ \frac{3\pi}{2}

Equations that reduce to tan x = k

When an equation has the form $a\sin x = b\cos x$ , dividing both sides by $\cos x$ gives $\tan x = \frac{b}{a}$ . This division is safe here: if $\cos x = 0$ then $\sin x = \pm 1$ , and the equation would read $\pm a = 0$ , which is false for $a \neq 0$ , so no solution is lost.

Exam-style questions on trigonometric equations

No calculator.

(a) Show that the equation $2\cos^2 x = 3\sin x$ can be written as $2\sin^2 x + 3\sin x - 2 = 0$ .

(b) Hence solve $2\cos^2 x = 3\sin x$ for $0 \le x \le 2\pi$ .

(c) Hence solve $2\cos^2 2x = 3\sin 2x$ for $0 \le x \le 2\pi$ .

Show the solution

(a) Using $\cos^2 x = 1 - \sin^2 x$ :

2(1 - \sin^2 x) = 3\sin x \quad \Rightarrow \quad 2 - 2\sin^2 x - 3\sin x = 0 \quad \Rightarrow \quad 2\sin^2 x + 3\sin x - 2 = 0

Every line is shown, because the answer is given.

(b) Factorise: $(2\sin x - 1)(\sin x + 2) = 0$ . Since $-1 \le \sin x \le 1$ , $\sin x = -2$ has no solutions. From $\sin x = \frac{1}{2}$ ,

x = \frac{\pi}{6} \quad \text{or} \quad x = \frac{5\pi}{6}

(c) "Hence" means use part (b). The equation is the same with $2x$ in place of $x$ , so $\sin 2x = \frac{1}{2}$ . For $0 \le x \le 2\pi$ , $0 \le 2x \le 4\pi$ :

2x = \frac{\pi}{6}, \ \frac{5\pi}{6}, \ \frac{13\pi}{6}, \ \frac{17\pi}{6} \quad \Rightarrow \quad x = \frac{\pi}{12}, \ \frac{5\pi}{12}, \ \frac{13\pi}{12}, \ \frac{17\pi}{12}

The second question needs a GDC for part (c), but parts (a), (b) and (d) can be done by hand.

GDC allowed. The depth of water, $d$ metres, at the entrance to a harbour $t$ hours after midnight is modelled by

d(t) = 1.8\cos(0.5t) + 4.2, \qquad 0 \le t \le 24

(a) Write down the greatest and least depths.

(b) Find the period of the model.

(c) A ship can enter the harbour only when the depth is more than $5$ m. Find the values of $t$ for which $d(t) = 5$ .

(d) Hence find the total time during the day when the ship can enter the harbour.

Show the solution

(a) Greatest depth $4.2 + 1.8 = 6$ m; least depth $4.2 - 1.8 = 2.4$ m.

(b) Period $= \frac{2\pi}{0.5} = 4\pi = 12.6$ hours (3 s.f.).

(c) $1.8\cos(0.5t) + 4.2 = 5$ gives $\cos(0.5t) = \frac{0.8}{1.8} = \frac{4}{9}$ . Let $u = 0.5t$ , so $0 \le u \le 12$ . The first solution is $u = \cos^{-1}\frac{4}{9} = 1.1102\ldots$ , and the others in the interval are $2\pi - 1.1102\ldots$ , $2\pi + 1.1102\ldots$ and $4\pi - 1.1102\ldots$ (the next one, $4\pi + 1.1102\ldots = 13.7$ , is beyond $12$ ). Doubling to return to $t$ :

t = 2.22, \quad 10.3, \quad 14.8, \quad 22.9 \quad (3 \text{ s.f.})

The same values come from the intersections of $y = d(t)$ and $y = 5$ on a GDC.

(d) The depth starts at its maximum ( $d(0) = 6$ ), so it is above $5$ m for $0 \le t < 2.2205$ , $10.346 < t < 14.787$ and $22.912 < t \le 24$ . The total time is

2.2205 + (14.787 - 10.346) + (24 - 22.912) = 7.75 \text{ hours (3 s.f.)}

A last type of question asks for the number of solutions rather than the solutions themselves.

Let $f(x) = 2\sin x + 1$ for $0 \le x \le 2\pi$ . Find the values of $k$ for which the equation $f(x) = k$ has exactly two solutions.

Show the solution

The graph of $f$ has amplitude $2$ and principal axis $y = 1$ . It starts at $(0, 1)$ , rises to a maximum $(\frac{\pi}{2}, 3)$ , falls through $(\pi, 1)$ to a minimum $(\frac{3\pi}{2}, -1)$ and ends at $(2\pi, 1)$ .

y = 2sin x + 1

Horizontal lines y = k meet the curve twice for −1 < k < 1 and 1 < k < 3, three times at k = 1 and once at k = 3 or k = −1.

A horizontal line $y = k$ meets the graph:

once if $k = 3$ or $k = -1$ (only the maximum or the minimum);
three times if $k = 1$ (at $x = 0$ , $\pi$ and $2\pi$ );
twice for every other $k$ between $-1$ and $3$ ;
not at all if $k > 3$ or $k < -1$ .

So there are exactly two solutions when $-1 < k < 1$ or $1 < k < 3$ .

Self-check

To do:
I can explain why a trigonometric equation usually has infinitely many solutions, and find all of them in a given interval.
To do:
I can find the second solution of $\sin x = k$ , $\cos x = k$ and $\tan x = k$ from the first, using the symmetry of the unit circle.
To do:
I can solve trigonometric equations exactly using exact values, and approximately using a calculator.
To do:
I can solve trigonometric equations graphically with a GDC.
To do:
I can solve equations such as $\sin 2x = 0.5$ or $\cos\left(x - \frac{\pi}{3}\right) = k$ by changing the interval first.
To do:
I can solve equations that are quadratic in $\sin x$ , $\cos x$ or $\tan x$ , and reject impossible values with a reason.
To do:
I can use the Pythagorean and double angle identities to rewrite an equation in terms of one function.
To do:
I can factorise instead of dividing by a function that could be zero.

Questions on 3.8

88 exam-style questions on this subtopic, written in the style of the IB papers. The first has its markscheme open. Counted on 20 October 2026

Question 1 · AA-000036 marks

Solve the equation $2\cos^2 x + \sin x = 1$ for $0 \le x \le 2\pi$ .

6 marks

Markscheme

Using $\cos^2 x = 1 - \sin^2 x$ : $2 - 2\sin^2 x + \sin x = 1$ M1

$2\sin^2 x - \sin x - 1 = 0$ A1

$(2\sin x + 1)(\sin x - 1) = 0$ , so $\sin x = -\tfrac{1}{2}$ or $\sin x = 1$ M1A1

$x = \frac{\pi}{2},\ \frac{7\pi}{6},\ \frac{11\pi}{6}$ A1A1

Award A1 for $\frac{\pi}{2}$ and A1 for both other values.

Question 2 · AA-001408 marks

A bead moves anticlockwise around a circular wire. The wire is modelled by the unit circle, centre O, with distances measured in metres. At time $t$ seconds, the position of the bead is

\left(\cos(0.8t), \sin(0.8t)\right), \quad 0 \le t \le 12.

(a)
Find the coordinates of the bead when $t = 2.5$ .
2 marks
(b)
Find the value of $t$ when the bead first returns to its starting position.
2 marks
(c)
Find the values of $t$ for which the $x$ -coordinate of the bead is $-0.6$ and its $y$ -coordinate is positive.
4 marks

Answer it, then compare with the markscheme.Check this one with a free account

Question 3 · AA-0014616 marks

A seat on an observation wheel starts at its lowest point. The height of the seat above the ground, $h$ metres, $t$ minutes later is modelled by

h(t) = 20 - 18\cos\left(\frac{\pi t}{12}\right), \quad 0 \le t \le 24.

(a)
Find the exact height of the seat when $t = 9$ .
2 marks
(b)
Write down the maximum height of the seat and the value of $t$ at which it occurs.
2 marks
(c)
Find the values of $t$ for which $h(t) = 29$ .
4 marks
(d)
Hence, find the length of time during one revolution for which the seat is more than 29 m above the ground.
2 marks
(e)
Find the rate at which the height of the seat is changing when $t = 2$ .
3 marks
(f)
Find the values of $t$ at which the seat is descending at a rate of $\dfrac{3\pi}{4}$ metres per minute.
3 marks

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One of 32 short multiple-choice questions on 3.8. Pick an answer to see why it's right or wrong.

How many solutions does $\sin^2 x = \frac14$ have for $0 \le x \le 2\pi$ ?

Key terms introduced here

Notes on 3.8 Solving trigonometric equations

What it means to solve a trigonometric equation

The value given by a calculator

Solving sin x = k, cos x = k and tan x = k

A method for every interval

Values with only one solution per cycle

Solving trigonometric equations graphically

Equations with a transformed angle

Equations from sinusoidal models

Equations leading to quadratic equations

Rejecting impossible values

Common factors: never divide by a function that could be zero

Using identities to get one function

The Pythagorean identity

The double angle identities

Equations that reduce to tan x = k

Exam-style questions on trigonometric equations

Self-check

Questions on 3.8

Markscheme

Try a quick-fire question

Carry on with 3.8 in Augment