Trigonometric Equations – OCR FSMQ Additional Maths

1 Why There Is More Than One Answer

PT5: solve simple trigonometric equations in given intervals. The phrase "in given intervals" is the whole difficulty — because trigonometric functions repeat, there is almost never just one solution.

Your calculator gives one solution only. $\sin^{-1}$, $\cos^{-1}$ and $\tan^{-1}$ each return a single "principal" value. Finding the rest is your job, and most of the marks are there.

The symmetry rules for a second solution

$\sin\theta = k$ → if $\theta_1$ works, so does $180° - \theta_1$
$\cos\theta = k$ → if $\theta_1$ works, so does $360° - \theta_1$
$\tan\theta = k$ → if $\theta_1$ works, so does $180° + \theta_1$

Then add or subtract whole periods

For $\sin$ and $\cos$: add or subtract $360°$ as needed
For $\tan$: add or subtract $180°$

Always sketch the curve and the horizontal line. Counting the crossings in the given interval tells you immediately how many answers to look for, which stops you stopping early.

2 The Basic Method

Rearrange until the equation reads $\sin\theta = k$, $\cos\theta = k$ or $\tan\theta = k$.
Find the principal value from the calculator.
Use the symmetry rule to find a second solution.
Add or subtract whole periods to reach every solution in the interval.
Discard anything outside the interval, and check nothing is missing against your sketch.

Worked Example 1 — A sine equation

Solve $\sin\theta = 0.4$ for $0° \leqslant \theta \leqslant 360°$.

①$\sin^{-1}(0.4) = 23.6°$ (1 d.p.)

②Second solution: $180° - 23.6° = 156.4°$

③Adding $360°$ takes us past the interval, so we stop.

④$\theta = 23.6°$ or $156.4°$

Worked Example 2 — A negative cosine

Solve $\cos\theta = -0.7$ for $0° \leqslant \theta \leqslant 360°$.

①$\cos^{-1}(-0.7) = 134.4°$ (1 d.p.) — the calculator handles the negative itself.

②Second solution: $360° - 134.4° = 225.6°$

③$\theta = 134.4°$ or $225.6°$

For cosine, the two solutions in $0°$–$360°$ are always symmetric about $180°$. Check: $134.4$ and $225.6$ average to $180$ ✓

Worked Example 3 — The specification's own example

Solve $\tan 2x = 0.5$ for $0° \leqslant x \leqslant 360°$.

①Work with the whole angle $2x$ first. As $x$ runs from $0°$ to $360°$, $2x$ runs from $0°$ to $720°$.

②$\tan^{-1}(0.5) = 26.565°$

③Tangent has period $180°$, so in $0°$ to $720°$: $2x = 26.565°,\; 206.565°,\; 386.565°,\; 566.565°$

④The next would be $746.565° > 720°$, so stop.

⑤Halve each: $x = 13.3°,\; 103.3°,\; 193.3°,\; 283.3°$ (1 d.p.)

Extend the interval before solving, not after. If you find only $x = 13.3°$ and then try to add $180°$, you will miss half the answers. Multiply the interval by the coefficient first.

3 When the Angle Is Not Just $\theta$

The substitution method

1. Let $u$ be the whole expression in the bracket
2. Convert the interval for $\theta$ into an interval for $u$
3. Find every solution for $u$ in that wider interval
4. Convert each one back to $\theta$

Worked Example 4 — A multiple angle

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

①Let $u = 3\theta$. As $\theta$ goes from $0°$ to $180°$, $u$ goes from $0°$ to $540°$.

②$\sin u = 0.5$: principal value $u = 30°$.

③Symmetry: $u = 180° - 30° = 150°$.

④Add $360°$ to each: $u = 390°$ and $u = 510°$. Both are under $540°$ ✓

⑤So $u = 30°,\; 150°,\; 390°,\; 510°$.

⑥Divide by $3$: $\theta = 10°,\; 50°,\; 130°,\; 170°$

Count the answers you expect. Multiplying the angle by $3$ triples the number of cycles, so expect roughly three times as many solutions.

Worked Example 5 — A shifted angle

Solve $\cos(\theta + 40°) = 0.6$ for $0° \leqslant \theta \leqslant 360°$.

①Let $u = \theta + 40°$. The interval becomes $40° \leqslant u \leqslant 400°$.

②$\cos^{-1}(0.6) = 53.13°$

③Symmetry: $u = 360° - 53.13° = 306.87°$

④Both lie in $40° \leqslant u \leqslant 400°$ ✓. Adding $360°$ gives $413.13°$, which is outside, so there are no more.

⑤So $u = 53.13°$ or $306.87°$.

⑥Subtract $40°$: $\theta = 13.1°$ or $266.9°$ (1 d.p.)

Check each $u$ against the shifted interval, not the original. A value of $u$ below $40°$ would give a negative $\theta$ and must be rejected.

4 Quadratic Trigonometric Equations

The strategy

Get everything in terms of one ratio, then treat it as a quadratic

Worked Example 6 — A quadratic in sine

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

①Let $s = \sin\theta$: $2s^2 - s - 1 = 0$

②$(2s+1)(s-1) = 0$, so $s = -\tfrac12$ or $s = 1$.

③$\sin\theta = -\tfrac12$: the related acute angle is $30°$, and sine is negative in the third and fourth quadrants.

④$\theta = 180° + 30° = 210°$ or $\theta = 360° - 30° = 330°$

⑤$\sin\theta = 1$: $\theta = 90°$ only (the maximum is reached once).

⑥$\theta = 90°,\; 210°,\; 330°$

Worked Example 7 — Needing an identity first

Solve $3\cos^2\theta + 5\sin\theta = 1$ for $0° \leqslant \theta \leqslant 360°$.

①Mixed ratios, so use $\cos^2\theta = 1 - \sin^2\theta$:

②$3\left(1-\sin^2\theta\right) + 5\sin\theta = 1$

③$3 - 3\sin^2\theta + 5\sin\theta = 1$

④$3\sin^2\theta - 5\sin\theta - 2 = 0$

⑤$(3\sin\theta + 1)(\sin\theta - 2) = 0$

⑥$\sin\theta = 2$ is impossible — reject it, and say so.

⑦$\sin\theta = -\tfrac13$: related acute angle $19.47°$; sine negative in the third and fourth quadrants.

⑧$\theta = 199.5°$ or $340.5°$ (1 d.p.)

State the rejection explicitly. Writing "$\sin\theta = 2$ has no solutions since $|\sin\theta| \leqslant 1$" earns a mark; silently ignoring it does not.

Worked Example 8 — Converting to a tangent

Solve $3\sin\theta = 2\cos\theta$ for $0° \leqslant \theta \leqslant 360°$.

①Divide both sides by $\cos\theta$:

②$3\tan\theta = 2$, so $\tan\theta = \tfrac23$.

③$\tan^{-1}\left(\tfrac23\right) = 33.69°$

④Tangent has period $180°$: $\theta = 33.69°$ or $213.69°$.

⑤$\theta = 33.7°$ or $213.7°$ (1 d.p.)

Dividing by $\cos\theta$ is safe here because if $\cos\theta$ were $0$ the equation would force $\sin\theta = 0$ too, and sine and cosine are never both zero. Where such a check is not automatic, treat $\cos\theta = 0$ as a separate case.

5 Quick Reference

Sketch first

Count the crossings to know how many answers.

Sine second

$180° - \theta_1$.

Cosine second

$360° - \theta_1$.

Tangent second

$180° + \theta_1$.

Periods

Add $360°$ ($\sin$, $\cos$) or $180°$ ($\tan$).

Multiple angle

Widen the interval before solving.

Shifted angle

Substitute $u$, shift the interval, solve, shift back.

Quadratics

Get one ratio only, then factorise.

Mixed ratios

Use $\sin^2 + \cos^2 \equiv 1$.

Reject clearly

Say why $\sin\theta = 2$ is impossible.

6 Practice Questions

Question 1

Solve $\sin\theta = 0.8$ for $0° \leqslant \theta \leqslant 360°$.

▶ Show solution

$\sin^{-1}(0.8) = 53.1°$

Second: $180° - 53.1° = 126.9°$

$\theta = 53.1°$ or $126.9°$ (1 d.p.)

Question 2

Solve $\cos\theta = 0.25$ for $0° \leqslant \theta \leqslant 360°$.

▶ Show solution

$\cos^{-1}(0.25) = 75.5°$

Second: $360° - 75.5° = 284.5°$

$\theta = 75.5°$ or $284.5°$ (1 d.p.)

Question 3

Solve $\tan\theta = 3$ for $0° \leqslant \theta \leqslant 360°$.

▶ Show solution

$\tan^{-1}(3) = 71.6°$

Adding the period $180°$: $251.6°$

$\theta = 71.6°$ or $251.6°$ (1 d.p.)

Question 4

Solve $\sin\theta = -0.3$ for $0° \leqslant \theta \leqslant 360°$.

▶ Show solution

Related acute angle: $\sin^{-1}(0.3) = 17.46°$.

Sine is negative in the third and fourth quadrants:

$180° + 17.46° = 197.5°$ and $360° - 17.46° = 342.5°$

$\theta = 197.5°$ or $342.5°$ (1 d.p.)

Question 5

Solve $2\cos\theta + 1 = 0$ for $0° \leqslant \theta \leqslant 360°$, giving exact answers.

▶ Show solution

$\cos\theta = -\dfrac12$

Related acute angle $60°$; cosine negative in the second and third quadrants.

$\theta = 120°$ or $240°$

Question 6

Solve $\sin 2\theta = 0.5$ for $0° \leqslant \theta \leqslant 360°$.

▶ Show solution

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

$\sin u = 0.5$: $u = 30°$, $150°$, and adding $360°$: $390°$, $510°$.

Halving: $\theta = 15°,\; 75°,\; 195°,\; 255°$

Question 7

Solve $\cos 3\theta = 0$ for $0° \leqslant \theta \leqslant 180°$.

▶ Show solution

Let $u = 3\theta$, so $0° \leqslant u \leqslant 540°$.

$\cos u = 0$ at $u = 90°$, $270°$, $450°$ — all within range.

$\theta = 30°,\; 90°,\; 150°$

Question 8

Solve $2\sin^2\theta = \sin\theta$ for $0° \leqslant \theta \leqslant 360°$.

▶ Show solution

Do not divide by $\sin\theta$ — that loses roots. Factorise instead:

$2\sin^2\theta - \sin\theta = 0 \;\Rightarrow\; \sin\theta\left(2\sin\theta - 1\right) = 0$

$\sin\theta = 0$: $\theta = 0°,\; 180°,\; 360°$

$\sin\theta = \tfrac12$: $\theta = 30°,\; 150°$

$\theta = 0°,\; 30°,\; 150°,\; 180°,\; 360°$

Question 9

Solve $5\sin\theta = 3\cos\theta$ for $0° \leqslant \theta \leqslant 360°$.

▶ Show solution

Dividing by $\cos\theta$: $\tan\theta = \dfrac{3}{5} = 0.6$

$\tan^{-1}(0.6) = 31.0°$

Adding $180°$: $211.0°$

$\theta = 31.0°$ or $211.0°$ (1 d.p.)

Question 10

The depth of water in a harbour, in metres, is modelled by $d = 7 + 3\sin(30t)°$, where $t$ is the time in hours after midnight.

(a) Find the depth at midnight and the maximum depth. (b) Find the period of the model and explain what it represents. (c) A boat needs a depth of at least $8.5$ m. Find, to the nearest minute, the first two times after midnight at which the depth is exactly $8.5$ m. (d) State the total time in each cycle during which the boat can enter.

▶ Show solution

(a) At $t = 0$: $d = 7 + 3\sin 0° = 7$ m.

The sine term is at most $1$, so the maximum depth is $7 + 3 = 10$ m.

(b) The angle is $30t$ degrees, so one full cycle needs $30t = 360$, giving $t = 12$ hours.

The period is $12$ hours, representing the time from one high tide to the next.

(c) Set $d = 8.5$:

$7 + 3\sin(30t)° = 8.5 \;\Rightarrow\; \sin(30t)° = 0.5$

Let $u = 30t$. For $0 \leqslant t \leqslant 12$ we have $0° \leqslant u \leqslant 360°$.

$\sin u = 0.5$ gives $u = 30°$ or $u = 150°$.

$t = \dfrac{30}{30} = 1$ hour, and $t = \dfrac{150}{30} = 5$ hours.

So the depth is $8.5$ m at 01:00 and 05:00.

(d) Between $t = 1$ and $t = 5$ the value of $\sin(30t)°$ exceeds $0.5$ — the sine curve is above the line $y = 0.5$ between those two crossings, peaking at $t = 3$ where $d = 10$ m.

So the depth is at least $8.5$ m for $5 - 1 = \mathbf{4}$ hours in each $12$-hour cycle.

Check at $t = 3$: $d = 7 + 3\sin 90° = 10 \geqslant 8.5$ ✓ and at $t = 6$: $d = 7 + 3\sin 180° = 7 < 8.5$ ✓

Trigonometric Equations (PT5) · OCR FSMQ Additional Maths · Created with MathJax