Solving trigonometric equations

AS · P1 · 12 min

A calculator gives one solution of $\sin x = 0.6$ , but the equation has infinitely many. Paper 1 asks for all the solutions in a stated interval, in degrees or radians, for equations such as $3\sin 2x + 1 = 0$ for $-\pi < x < \pi$ and $3\sin^2\theta - 5\cos\theta - 1 = 0$ for $0^\circ \le \theta \le 360^\circ$ (both examples from the syllabus itself). The method is always the same: get one function equal to a number, find the principal value, use the symmetry of the graph for the second solution, and use the period to collect the rest. General solutions are not required.

One principal value, then symmetry

The principal value from $\sin^{-1}$ , $\cos^{-1}$ or $\tan^{-1}$ is the first solution. The others come from the graphs:

Sine is symmetrical about $90^\circ$ , so if $\alpha$ is a solution, so is $180^\circ - \alpha$ (or $\pi - \alpha$ ).
Cosine is symmetrical about $0^\circ$ (and $360^\circ$ ), so if $\alpha$ is a solution, so is $-\alpha$ , and hence $360^\circ - \alpha$ (or $2\pi - \alpha$ ).
Tangent has period $180^\circ$ , so if $\alpha$ is a solution, so is $\alpha + 180^\circ$ (or $\alpha + \pi$ ).

Then add or subtract the period ( $360^\circ$ for sine and cosine, $180^\circ$ for tangent) to reach every solution in the interval.

y = sin(x*pi/180) y = 0.6

$\sin x = 0.6$ for $0^\circ \le x \le 360^\circ$ : the principal value is $36.9^\circ$ , and the symmetry about $90^\circ$ gives $180^\circ - 36.9^\circ = 143.1^\circ$ . There are no others in the interval.

Key result

For the principal value $\alpha$ :

Equation	Second solution in the same cycle	Then add or subtract
$\sin x = k$	$180^\circ - \alpha$ or $\pi - \alpha$	$360^\circ$ or $2\pi$
$\cos x = k$	$360^\circ - \alpha$ (or $-\alpha$ ), or $2\pi - \alpha$	$360^\circ$ or $2\pi$
$\tan x = k$	$\alpha + 180^\circ$ or $\alpha + \pi$	$180^\circ$ or $\pi$

There are no solutions of $\sin x = k$ or $\cos x = k$ if $|k| > 1$ .

The CAST diagram from Exact values and angles of any size gives the same answers: find the related acute angle and place it in each quadrant where the function has the right sign. Use whichever you find more reliable, and check with a sketch.

Multiple and shifted angles

For $\sin 2x = k$ with $0^\circ \le x \le 360^\circ$ , let $u = 2x$ . Then $u$ runs over $0^\circ \le u \le 720^\circ$ , two full cycles, so there are twice as many solutions. Find every $u$ in the new interval, then divide by $2$ .

The same substitution handles a shifted angle: for $\cos(x + 30^\circ) = k$ with $0^\circ \le x \le 360^\circ$ , let $u = x + 30^\circ$ , so $30^\circ \le u \le 390^\circ$ .

Solving a trigonometric equation

Rearrange so a single trigonometric function of a single angle equals a number. Use identities if the equation mixes functions, and factorise if it is a product or a quadratic.
If the angle is $bx + c$ , substitute $u = bx + c$ and transform the interval: apply the same operations to its ends.
Find the principal value $\alpha = \sin^{-1}k$ , $\cos^{-1}k$ or $\tan^{-1}k$ .
Find the second value from the symmetry, then add and subtract periods until you have every value of $u$ in the transformed interval.
Convert back to $x$ , and list the solutions in order, to the accuracy asked.
Check the count against a sketch, and reject any value outside the interval.

Equations that need rearranging

Two functions of the same angle

$a\sin x = b\cos x$ : divide by $\cos x$ to get $\tan x = \tfrac{b}{a}$ . This is safe because $\cos x = 0$ would force $\sin x = 0$ too, which never happens at the same time.
A product equal to zero, such as $\cos x(2\sin x - 1) = 0$ : each factor gives its own equation. Never divide by a function that might be zero: that loses solutions.

Quadratics in a trigonometric function

If the equation contains $\sin^2 x$ and $\cos x$ (or $\cos^2 x$ and $\sin x$ ), use $\sin^2 x + \cos^2 x = 1$ to change the square, so only one function remains (see Trigonometric identities). Then factorise or use the formula, as in Equations that are quadratic in a function of x. Reject any root outside $-1 \le k \le 1$ for sine and cosine. Equations with $\tan x$ and $\dfrac{1}{\tan x}$ become quadratics in $\tan x$ after multiplying through by $\tan x$ .

Squares

$\tan^2 x = 3$ means $\tan x = \sqrt{3}$ or $\tan x = -\sqrt{3}$ . Both signs give solutions.

Worked examples

A negative value of sine

Solve $\sin x = -0.6$ for $0^\circ \le x \le 360^\circ$ .

Solution

$\alpha = \sin^{-1}(-0.6) = -36.87^\circ$ , which is outside the interval.

The second value from the symmetry is $180^\circ - (-36.87^\circ) = 216.87^\circ$ . Adding $360^\circ$ to $\alpha$ gives $323.13^\circ$ .

x = 216.9^\circ,\ 323.1^\circ

(Both are where sine is negative: the third and fourth quadrants.)

A double angle in degrees

Solve $\cos 2x = 0.4$ for $0^\circ \le x \le 360^\circ$ .

Solution

Let $u = 2x$ , so $0^\circ \le u \le 720^\circ$ .

$\cos^{-1}0.4 = 66.42^\circ$ ; the second value is $360^\circ - 66.42^\circ = 293.58^\circ$ . Adding $360^\circ$ to each: $426.42^\circ$ and $653.58^\circ$ .

Dividing by $2$ :

x = 33.2^\circ,\ 146.8^\circ,\ 213.2^\circ,\ 326.8^\circ

y = cos(2x*pi/180) y = 0.4

The sketch confirms four solutions: $y = \cos 2x$ makes two complete waves between $0^\circ$ and $360^\circ$ , and the line $y = 0.4$ cuts each wave twice.

The syllabus example in radians

Solve $3\sin 2x + 1 = 0$ for $-\pi < x < \pi$ .

Solution

$\sin 2x = -\tfrac{1}{3}$ . Let $u = 2x$ , so $-2\pi < u < 2\pi$ (two full cycles, so expect four solutions).

$\alpha = \sin^{-1}\left(-\tfrac{1}{3}\right) = -0.3398$ . The second value is $\pi - (-0.3398) = 3.4814$ .

Adding or subtracting $2\pi$ to stay in $(-2\pi, 2\pi)$ : $-0.3398 + 2\pi = 5.9434$ and $3.4814 - 2\pi = -2.8018$ .

So $u = -2.8018,\ -0.3398,\ 3.4814,\ 5.9434$ , and $x = \tfrac{u}{2}$ :

x = -1.40,\ -0.170,\ 1.74,\ 2.97

A shifted angle with an exact answer

Solve $\sin\left(2x - \tfrac{\pi}{6}\right) = \tfrac{1}{2}$ for $0 \le x \le \pi$ .

Solution

Let $u = 2x - \tfrac{\pi}{6}$ . Then $-\tfrac{\pi}{6} \le u \le \tfrac{11\pi}{6}$ .

$\sin u = \tfrac{1}{2}$ gives $u = \tfrac{\pi}{6}$ and $u = \pi - \tfrac{\pi}{6} = \tfrac{5\pi}{6}$ . The next ones, $\tfrac{13\pi}{6}$ and $-\tfrac{7\pi}{6}$ , are outside the interval.

$2x = u + \tfrac{\pi}{6}$ gives $2x = \tfrac{\pi}{3}$ or $\pi$ , so

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

The syllabus example with an identity

Solve $3\sin^2\theta - 5\cos\theta - 1 = 0$ for $0^\circ \le \theta \le 360^\circ$ .

Solution

Replace $\sin^2\theta$ by $1 - \cos^2\theta$ :

\begin{aligned} 3\left(1 - \cos^2\theta\right) - 5\cos\theta - 1 &= 0 \\ 3\cos^2\theta + 5\cos\theta - 2 &= 0 \\ (3\cos\theta - 1)(\cos\theta + 2) &= 0 \end{aligned}

$\cos\theta = -2$ has no solutions. $\cos\theta = \tfrac{1}{3}$ gives $\theta = 70.5^\circ$ or $360^\circ - 70.5^\circ = 289.5^\circ$ .

Sine and cosine together

Solve $3\sin x = 5\cos x$ for $0^\circ \le x \le 360^\circ$ .

Solution

Divide by $\cos x$ (it cannot be zero at a solution, because then $\sin x$ would be zero too): $\tan x = \tfrac{5}{3}$ .

$\tan^{-1}\tfrac{5}{3} = 59.04^\circ$ , and tangent repeats every $180^\circ$ :

x = 59.0^\circ,\ 239.0^\circ

Factorising instead of dividing

Solve $2\sin x\cos x = \cos x$ for $0^\circ \le x \le 360^\circ$ .

Solution

Do not divide by $\cos x$ . Rearrange and factorise:

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

$\cos x = 0$ : $x = 90^\circ, 270^\circ$ .

$\sin x = \tfrac{1}{2}$ : $x = 30^\circ, 150^\circ$ .

x = 30^\circ,\ 90^\circ,\ 150^\circ,\ 270^\circ

Dividing by $\cos x$ at the start would have lost $90^\circ$ and $270^\circ$ .

A quadratic in a double angle

Solve $2\cos^2 2x + 3\sin 2x - 3 = 0$ for $0^\circ \le x \le 180^\circ$ .

Solution

Replace $\cos^2 2x$ by $1 - \sin^2 2x$ :

\begin{aligned} 2 - 2\sin^2 2x + 3\sin 2x - 3 &= 0 \\ 2\sin^2 2x - 3\sin 2x + 1 &= 0 \\ (2\sin 2x - 1)(\sin 2x - 1) &= 0 \end{aligned}

Let $u = 2x$ , with $0^\circ \le u \le 360^\circ$ .

$\sin u = \tfrac{1}{2}$ : $u = 30^\circ, 150^\circ$ . $\sin u = 1$ : $u = 90^\circ$ .

Dividing by $2$ :

x = 15^\circ,\ 45^\circ,\ 75^\circ

Watch out

Stopping at the calculator value. $\sin^{-1}(-0.6) = -36.9^\circ$ is not even in $0^\circ$ to $360^\circ$ . Always generate the full set.

Not changing the interval for $2x$ . For $\cos 2x = 0.4$ on $0^\circ \le x \le 360^\circ$ you need $u$ up to $720^\circ$ . Working only to $360^\circ$ finds half the solutions.

Dividing by $\sin x$ or $\cos x$ . This loses the solutions where that function is zero. Factorise instead. (Dividing $a\sin x = b\cos x$ by $\cos x$ is the one safe exception.)

Keeping impossible roots. $\cos\theta = -2$ or $\sin\theta = 1.5$ give no solutions; say so and move on.

Mixing degrees and radians. If the interval is in radians, give radian answers and use radian mode.

Forgetting the negative root. $\tan^2 x = 3$ gives $\tan x = \pm\sqrt{3}$ .

Exam tip

Accuracy. Degrees to $1$ decimal place, radians to $3$ significant figures, unless stated otherwise. Exact answers (such as $\tfrac{\pi}{6}$ ) are expected when the values are standard.
Extra solutions are penalised. An answer outside the interval, or a spurious solution from an impossible root, typically loses the final A mark even if all correct solutions are present.
Marks. A typical 4-mark question: M1 for using the identity correctly, A1 for the correct quadratic, M1 for solving and using the inverse, A1 for all solutions in the interval and no others.
Show the principal value and the method for the second value, such as " $180^\circ - 36.9^\circ$ ". If you slip, the method mark is still available.
"Hence". If you have just proved an identity, use it: replace the complicated side and solve the simpler equation.
Interval ends. Check whether the interval uses $<$ or $\le$ ; a solution exactly at an end may or may not be included.

Summary

Rearrange to one function of one angle equal to a number; use identities and factorising first.
Principal value from the calculator, then: sine $180^\circ - \alpha$ , cosine $360^\circ - \alpha$ , tangent $\alpha + 180^\circ$ .
Add or subtract the period ( $360^\circ$ or $2\pi$ ; $180^\circ$ or $\pi$ for tangent) to cover the interval.
For $\sin(bx + c)$ and similar, substitute $u = bx + c$ , transform the interval, solve for $u$ , then convert back.
$a\sin x = b\cos x$ becomes $\tan x = \tfrac{b}{a}$ ; products equal to zero split into separate equations.
Quadratics: use $\sin^2 + \cos^2 = 1$ , solve, reject values outside $[-1, 1]$ .
List solutions in order, to the accuracy asked, and check the number against a sketch.

Practice questions

Question

Solve $\tan x = -2$ for $-180^\circ \le x \le 180^\circ$ .
Solve $\sin 3x = \tfrac{1}{2}$ for $0^\circ \le x \le 180^\circ$ .
Solve $2\cos x + \sqrt{3} = 0$ for $0 \le x \le 2\pi$ , giving exact answers.
Solve $\cos(x + 60^\circ) = -\tfrac{1}{2}$ for $0^\circ \le x \le 360^\circ$ .
Solve $2\cos^2 x + \sin x = 1$ for $0^\circ \le x \le 360^\circ$ .
Solve $\sin\theta\cos\theta = \sin\theta$ for $0 \le \theta \le 2\pi$ .
Solve $5\sin x = 2\cos x$ for $0^\circ \le x \le 360^\circ$ .
Solve $\tan x + \dfrac{2}{\tan x} = 3$ for $0^\circ \le x \le 360^\circ$ .
Solve $4\sin^2 2\theta = 3$ for $0 \le \theta \le \pi$ , giving exact answers.
(a) Show that the equation $3\sin x\tan x = 8$ can be written as $3\cos^2 x + 8\cos x - 3 = 0$ . (b) Hence solve $3\sin 2y\tan 2y = 8$ for $0^\circ \le y \le 180^\circ$ .

Answers

$\tan^{-1}(-2) = -63.4^\circ$ . Adding $180^\circ$ : $116.6^\circ$ . So $x = -63.4^\circ, 116.6^\circ$ .
$u = 3x$ , $0^\circ \le u \le 540^\circ$ . $\sin u = \tfrac{1}{2}$ : $u = 30^\circ, 150^\circ, 390^\circ, 510^\circ$ . So $x = 10^\circ, 50^\circ, 130^\circ, 170^\circ$ .
$\cos x = -\tfrac{\sqrt{3}}{2}$ . The related angle is $\tfrac{\pi}{6}$ , and cosine is negative in the second and third quadrants: $x = \tfrac{5\pi}{6}, \tfrac{7\pi}{6}$ .
$u = x + 60^\circ$ , $60^\circ \le u \le 420^\circ$ . $\cos u = -\tfrac{1}{2}$ : $u = 120^\circ, 240^\circ$ (the next, $480^\circ$ , is too large). So $x = 60^\circ, 180^\circ$ .
$2(1 - \sin^2 x) + \sin x - 1 = 0$ , so $2\sin^2 x - \sin x - 1 = 0$ , $(2\sin x + 1)(\sin x - 1) = 0$ . $\sin x = 1$ : $x = 90^\circ$ . $\sin x = -\tfrac{1}{2}$ : $x = 210^\circ, 330^\circ$ . Solutions $90^\circ, 210^\circ, 330^\circ$ .
$\sin\theta(\cos\theta - 1) = 0$ . $\sin\theta = 0$ : $\theta = 0, \pi, 2\pi$ . $\cos\theta = 1$ : $\theta = 0, 2\pi$ . Solutions $0, \pi, 2\pi$ .
$\tan x = \tfrac{2}{5}$ : $x = 21.8^\circ$ and $201.8^\circ$ .
Multiply by $\tan x$ : $\tan^2 x - 3\tan x + 2 = 0$ , $(\tan x - 1)(\tan x - 2) = 0$ . $\tan x = 1$ : $45^\circ, 225^\circ$ . $\tan x = 2$ : $63.4^\circ, 243.4^\circ$ . Solutions $45^\circ, 63.4^\circ, 225^\circ, 243.4^\circ$ .
$\sin 2\theta = \pm\tfrac{\sqrt{3}}{2}$ . With $u = 2\theta$ , $0 \le u \le 2\pi$ : $u = \tfrac{\pi}{3}, \tfrac{2\pi}{3}$ (positive) and $\tfrac{4\pi}{3}, \tfrac{5\pi}{3}$ (negative). So $\theta = \tfrac{\pi}{6}, \tfrac{\pi}{3}, \tfrac{2\pi}{3}, \tfrac{5\pi}{6}$ .
(a) $3\sin x \cdot \dfrac{\sin x}{\cos x} = 8$ , so $3\sin^2 x = 8\cos x$ , $3(1 - \cos^2 x) = 8\cos x$ , and $3\cos^2 x + 8\cos x - 3 = 0$ . (b) With $x = 2y$ : $(3\cos 2y - 1)(\cos 2y + 3) = 0$ . $\cos 2y = -3$ is impossible, so $\cos 2y = \tfrac{1}{3}$ . For $0^\circ \le 2y \le 360^\circ$ : $2y = 70.53^\circ$ or $289.47^\circ$ , so $y = 35.3^\circ$ or $144.7^\circ$ .

One principal value, then symmetry

Multiple and shifted angles

Equations that need rearranging

Two functions of the same angle

Quadratics in a trigonometric function

Squares

Worked examples

Practice questions

Console