Trigonometric Equations – A Level Maths Year 1

1 Why Trig Equations Have Many Solutions

The equation $\sin x = 0.5$ has infinitely many solutions, because the sine graph keeps repeating. A question therefore always gives an interval, and you must find every solution in it — no more, no fewer. This page is part of DfE content area E · Trigonometry; in Year 1 all angles are in degrees.

The principal value is what the calculator gives: $\sin^{-1}$ and $\tan^{-1}$ return a value $\alpha$ with $-90° \leqslant \alpha \leqslant 90°$, and $\cos^{-1}$ returns a value with $0° \leqslant \alpha \leqslant 180°$. Every other solution is found from it using the symmetry and period of the graph.

The second solution in one period

$\sin x = k$: $x = \alpha$ and $180° - \alpha$ · $\cos x = k$: $x = \alpha$ and $360° - \alpha$ (or $-\alpha$) · $\tan x = k$: $x = \alpha$ and $\alpha + 180°$
then add or subtract $360°$ (sin, cos) or $180°$ (tan) to reach every solution in the interval

2 Solving $\sin x = k$, $\cos x = k$ and $\tan x = k$

Worked Example 1 — Exact answers

Solve $\sin x = \tfrac12$ for $0° \leqslant x \leqslant 360°$.

①Principal value: $x = 30°$.

②Second value: $180° - 30° = 150°$.

③Adding $360°$ takes both out of range. Solutions: $x = 30°, 150°$ (the two red points above).

Worked Example 2 — A negative cosine

Solve $\cos x = -0.3$ for $0° \leqslant x \leqslant 360°$, giving answers to 1 d.p.

①$x = \cos^{-1}(-0.3) = 107.457\ldots°$

②Second value: $360° - 107.46° = 252.54°$

③$x = 107.5°, 252.5°$

Worked Example 3 — Tan, on a symmetric interval

Solve $\tan x = 2$ for $-180° \leqslant x \leqslant 180°$.

①$x = \tan^{-1} 2 = 63.43\ldots°$

②Tan repeats every $180°$: $63.43° - 180° = -116.57°$. Adding $180°$ instead gives $243.4°$, which is out of range.

③$x = -116.6°, 63.4°$

Worked Example 4 — A negative principal value

Solve $\sin x = -0.6$ for $-180° \leqslant x \leqslant 180°$.

①$x = \sin^{-1}(-0.6) = -36.87°$, which is in range.

②Second value: $180° - (-36.87°) = 216.87°$ — out of range, so subtract $360°$: $-143.13°$.

③$x = -143.1°, -36.9°$

Sense check with the graph: sine is negative between $-180°$ and $0°$, so both answers must be negative ✓

3 Equations of the Form $\sin(n\theta + \alpha) = k$

When the angle inside the function is not just $x$, solve for the whole angle first, over a correspondingly transformed interval, and only then rearrange.

Let $u = n\theta + \alpha$. Transform the interval: if $a \leqslant \theta \leqslant b$, then $na + \alpha \leqslant u \leqslant nb + \alpha$.
Find every solution for $u$ in the new interval.
Convert each one back: $\theta = \dfrac{u - \alpha}{n}$.

Worked Example 5 — A multiple angle

Solve $\sin 2x = \dfrac{\sqrt3}{2}$ for $0° \leqslant x \leqslant 360°$.

①Let $u = 2x$, so $0° \leqslant u \leqslant 720°$.

②$\sin u = \tfrac{\sqrt3}{2}$: $u = 60°, 120°$, and adding $360°$: $420°, 480°$.

③$x = \tfrac{u}{2}$: $x = 30°, 60°, 210°, 240°$.

Doubling the angle doubles the number of solutions — exactly as the graph of $y = \sin 2x$ (two cycles in $360°$) predicts.

Worked Example 6 — A shifted angle

Solve $\cos(x - 30°) = 0.5$ for $0° \leqslant x \leqslant 360°$.

①Let $u = x - 30°$, so $-30° \leqslant u \leqslant 330°$.

②$\cos u = 0.5$: $u = 60°$ and $u = 360° - 60° = 300°$. ($u = -60°$ is outside the interval.)

③$x = u + 30°$: $x = 90°, 330°$.

Worked Example 7 — When the principal value is out of range

Solve $\sin(2\theta + 30°) = 0.4$ for $0° \leqslant \theta \leqslant 180°$.

①Let $u = 2\theta + 30°$, so $30° \leqslant u \leqslant 390°$.

②Principal value $\sin^{-1} 0.4 = 23.58°$ — not in the interval.

③Other values: $180° - 23.58° = 156.42°$ ✓ and $23.58° + 360° = 383.58°$ ✓

④$\theta = \dfrac{u - 30°}{2}$: $\theta = 63.2°, 176.8°$ (1 d.p.)

Students who start from the calculator value and only "go forwards" lose the solution at $383.58°$, or wrongly keep $23.58°$. Always list candidates and check each against the transformed interval.

4 Quadratic Equations in $\sin$, $\cos$ or $\tan$

Treat $\sin x$ (or $\cos x$, $\tan x$) as the unknown in an ordinary quadratic. Factorise, then solve each factor as a basic equation. If the equation mixes functions, use an identity from Trigonometric Identities first.

Worked Example 8 — Factorising

Solve $2\sin^2 x - \sin x - 1 = 0$ for $0° \leqslant x \leqslant 360°$.

①Think $2s^2 - s - 1 = (2s + 1)(s - 1)$: so $(2\sin x + 1)(\sin x - 1) = 0$.

②$\sin x = 1$: $x = 90°$.

③$\sin x = -\tfrac12$: principal $-30°$ (out of range); solutions $180° + 30° = 210°$ and $360° - 30° = 330°$.

④$x = 90°, 210°, 330°$

Worked Example 9 — Using $\sin^2 x + \cos^2 x = 1$

Solve $2\cos^2 x + 3\sin x = 3$ for $0° \leqslant x \leqslant 360°$.

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

②$(2\sin x - 1)(\sin x - 1) = 0$

③$\sin x = \tfrac12$: $x = 30°, 150°$. $\sin x = 1$: $x = 90°$.

④$x = 30°, 90°, 150°$

Worked Example 10 — Using $\tan x = \dfrac{\sin x}{\cos x}$

Solve $3\sin x = 4\cos x$ for $0° \leqslant x \leqslant 360°$.

①Divide by $\cos x$: $\tan x = \tfrac43$. (No solutions are lost: if $\cos x = 0$ then $\sin x = \pm1$ and the equation would say $\pm3 = 0$.)

②$x = 53.1°$ and $53.13° + 180° = 233.1°$.

Worked Example 11 — Don't divide away solutions

Solve $2\sin x\cos x = \sin x$ for $0° \leqslant x \leqslant 360°$.

①Do not cancel $\sin x$ — it could be zero. Instead: $\sin x(2\cos x - 1) = 0$.

②$\sin x = 0$: $x = 0°, 180°, 360°$.

③$\cos x = \tfrac12$: $x = 60°, 300°$.

④$x = 0°, 60°, 180°, 300°, 360°$

Worked Example 12 — Square roots give $\pm$

Solve $\tan^2 x = 3$ for $0° \leqslant x \leqslant 360°$.

①$\tan x = \sqrt3$ or $\tan x = -\sqrt3$.

②$\tan x = \sqrt3$: $x = 60°, 240°$. $\tan x = -\sqrt3$: $x = 120°, 300°$.

③$x = 60°, 120°, 240°, 300°$

5 Quick Reference

sin

$\alpha$, $180° - \alpha$, then $\pm 360°$.

cos

$\alpha$, $360° - \alpha$, then $\pm 360°$.

tan

$\alpha$, then $\pm 180°$.

Transformed angle

Transform the interval first, solve, then convert back.

$\sin nx$

$n$ times as many solutions.

Quadratic

Factorise in $\sin x$; reject any value $k$ with $k < -1$ or $k > 1$.

Mixed sin and cos²

Use $\cos^2 x = 1 - \sin^2 x$.

$a\sin x = b\cos x$

$\tan x = \dfrac{b}{a}$.

Never cancel

Factorise out a common $\sin x$ or $\cos x$.

Square root

Remember the negative root.

6 Practice Questions

Question 1

Solve $\cos x = 0.6$ for $0° \leqslant x \leqslant 360°$, giving answers to 1 d.p.

▶ Show solution

$x = \cos^{-1} 0.6 = 53.13°$, and $360° - 53.13° = 306.87°$.

$x = 53.1°, 306.9°$

Question 2

Solve $\tan x = -1$ for $0° \leqslant x \leqslant 360°$.

▶ Show solution

Principal value $-45°$, out of range.

Add $180°$: $135°$. Add another $180°$: $315°$.

$x = 135°, 315°$

Question 3

Find the exact solutions of $\sin x = -\dfrac{\sqrt3}{2}$ for $0° \leqslant x \leqslant 360°$.

▶ Show solution

Reference angle $60°$. Sine is negative in the third and fourth quadrants.

$x = 180° + 60° = 240°$ and $x = 360° - 60° = 300°$.

Question 4

Solve $\sin 3x = 0.5$ for $0° \leqslant x \leqslant 180°$.

▶ Show solution

$u = 3x$, $0° \leqslant u \leqslant 540°$.

$\sin u = 0.5$: $u = 30°, 150°, 390°, 510°$.

$x = 10°, 50°, 130°, 170°$

Question 5

Solve $\tan(x + 45°) = \sqrt3$ for $0° \leqslant x \leqslant 360°$.

▶ Show solution

$u = x + 45°$, $45° \leqslant u \leqslant 405°$.

$\tan u = \sqrt3$: $u = 60°, 240°$ ($420°$ is out of range).

$x = 15°, 195°$

Question 6

Solve $5\sin x = 2\cos x$ for $0° \leqslant x \leqslant 360°$.

▶ Show solution

$\tan x = \tfrac25 = 0.4$

$x = 21.8°$ and $21.80° + 180° = 201.8°$.

Question 7

Solve $2\cos^2 x = \cos x$ for $0° \leqslant x \leqslant 360°$.

▶ Show solution

$2\cos^2 x - \cos x = 0 \;\Rightarrow\; \cos x(2\cos x - 1) = 0$

$\cos x = 0$: $x = 90°, 270°$.

$\cos x = \tfrac12$: $x = 60°, 300°$.

$x = 60°, 90°, 270°, 300°$

Question 8

In this question you must show detailed reasoning. Solve $6\cos^2 x + \sin x - 5 = 0$ for $0° \leqslant x \leqslant 360°$.

▶ Show solution

$6(1 - \sin^2 x) + \sin x - 5 = 0 \;\Rightarrow\; 6\sin^2 x - \sin x - 1 = 0$

$(3\sin x + 1)(2\sin x - 1) = 0$

$\sin x = \tfrac12$: $x = 30°, 150°$.

$\sin x = -\tfrac13$: principal $-19.47°$; solutions $180° + 19.47° = 199.47°$ and $360° - 19.47° = 340.53°$.

$x = 30°, 150°, 199.5°, 340.5°$

Question 9

Solve $2\cos(2\theta - 40°) = -1$ for $0° \leqslant \theta \leqslant 180°$.

▶ Show solution

$\cos(2\theta - 40°) = -\tfrac12$. Let $u = 2\theta - 40°$, so $-40° \leqslant u \leqslant 320°$.

$\cos u = -\tfrac12$: $u = 120°, 240°$ (and $-120°$, which is out of range).

$\theta = \dfrac{u + 40°}{2}$: $\theta = 80°, 140°$.

Question 10

The depth of water $d$ metres in a harbour $t$ hours after midnight is modelled by $d = 5 + 3\sin(30t)°$ for $0 \leqslant t \leqslant 24$.

(a) Find the maximum depth and the first time it occurs. (b) A boat can only use the harbour when the depth is at least $6.5$ m. Find the times during the day when it can do so. (c) For how many hours in total is this? (d) Find the first time the depth is $3.5$ m.

▶ Show solution

(a) Maximum $5 + 3 = 8$ m when $\sin(30t)° = 1$, i.e. $30t = 90$, $t = 3$: at 03:00.

(b) $5 + 3\sin(30t)° \geqslant 6.5 \;\Rightarrow\; \sin(30t)° \geqslant \tfrac12$.

Let $u = 30t$, $0 \leqslant u \leqslant 720$. $\sin u = \tfrac12$ at $u = 30, 150, 390, 510$, and $\sin u \geqslant \tfrac12$ between each pair.

$30 \leqslant u \leqslant 150$ or $390 \leqslant u \leqslant 510$, so $1 \leqslant t \leqslant 5$ or $13 \leqslant t \leqslant 17$.

From 01:00 to 05:00 and from 13:00 to 17:00.

(c) $4 + 4 = 8$ hours.

(d) $\sin(30t)° = -\tfrac12$: the first $u$ in range is $180 + 30 = 210$, so $t = 7$: at 07:00.

Trigonometric Equations · A Level Maths Year 1 · Created with MathJax