Match the interval in the question. If it involves, work and answer in radians.

1.05PureStretch

Trigonometric Equations

Q: How do I know how many solutions to expect?

Sketch the graph over the interval and draw the line y = k. Each intersection is one solution.

Q: Why can’t I divide by?

If = 0 is a solution, dividing by it removes that solution. Factorising keeps it.

Q: What if the principal value is negative?

Add a full period to bring it into the interval, and use symmetry to find the partner solution.

Your calculator gives exactly one answer to $\sin\theta = 0.4$ . But the sine graph repeats forever, so the equation has infinitely many solutions — and an exam question asking for "all solutions in $0^{\circ} \le \theta < 360^{\circ}$ " wants every one of them in that interval. The method is always the same: find the principal value, then use symmetry and period to generate the rest.

45 min Trigonometry

The big picture

Solving trigonometric equations combines everything in this chapter. Ratios and their graphs tell you how many solutions to expect and where they sit. Identities turn an equation with mixed functions into one with a single function, often a quadratic. And transformations explain what happens with multiple angles: $\sin 2\theta = \tfrac12$ has twice as many solutions as $\sin\theta = \tfrac12$ in the same interval, because the graph is squashed. The disciplined technique — adjust the interval for the argument, list all values, then transform back — is what separates full marks from answers that stop at the calculator value.

What you'll be able to do

Find all solutions of $\sin\theta = k$ , $\cos\theta = k$ and $\tan\theta = k$ in a given interval
Solve equations with multiple or shifted angles such as $\cos(2\theta - 30^{\circ}) = k$
Solve quadratic equations in a trigonometric function
Use identities to reduce equations to a single function, working in degrees or radians

Principal values and the other solutions

The inverse functions on your calculator return one $\textbf{principal value}$ : $\sin^{-1}$ and $\tan^{-1}$ give values in $[-90^{\circ}, 90^{\circ}]$ , and $\cos^{-1}$ gives values in $[0^{\circ}, 180^{\circ}]$ .

The second solution in each $360^{\circ}$ cycle comes from symmetry: if $\sin\theta = k$ has solution $\alpha$ , the other is $180^{\circ} - \alpha$ ; for $\cos\theta = k$ the other is $360^{\circ} - \alpha$ (equivalently $-\alpha$ ); for $\tan\theta = k$ solutions repeat every $180^{\circ}$ .

Then add or subtract whole periods ( $360^{\circ}$ for sine and cosine, $180^{\circ}$ for tangent) until you have every solution in the interval.

\sin\theta = k: \ \theta = \alpha,\ 180^{\circ} - \alpha \ (+360^{\circ}n)

Sine solutions.

\cos\theta = k: \ \theta = \pm\alpha \ (+360^{\circ}n)

Cosine solutions.

\tan\theta = k: \ \theta = \alpha \ (+180^{\circ}n)

Tangent solutions.

Example 1Solve

\cos\theta = -0.3

for

0^{\circ} \le \theta < 360^{\circ}

1Principal value:

\cos^{-1}(-0.3) = 107.46^{\circ}

2Other solution:

360^{\circ} - 107.46^{\circ} = 252.54^{\circ}

Answer

\theta = 107.5^{\circ}

252.5^{\circ}

Tip — Sketch the graph with a horizontal line at $y = k$ before listing solutions. The number of intersections in the interval tells you how many answers to find.

Multiple and shifted angles

For $\sin(2\theta) = k$ with $0^{\circ} \le \theta < 360^{\circ}$ , substitute $x = 2\theta$ . The interval for $x$ becomes $0^{\circ} \le x < 720^{\circ}$ — two full cycles, so up to four solutions.

Find all $x$ in the new interval, then convert back: $\theta = \tfrac{x}{2}$ .

For a shifted angle like $\theta - 40^{\circ}$ , shift the interval by the same amount before solving.

Example 1Solve

\tan(3\theta) = 1

for

0^{\circ} \le \theta < 180^{\circ}

1Let

x = 3\theta

, so

0^{\circ} \le x < 540^{\circ}

2Principal value

x = 45^{\circ}

; add

180^{\circ}

repeatedly:

45^{\circ}, 225^{\circ}, 405^{\circ}

3The next,

585^{\circ}

, is outside the interval. Divide by 3.

Answer

\theta = 15^{\circ}, 75^{\circ}, 135^{\circ}

Example 2Solve

\sin\left(\theta + \tfrac{\pi}{6}\right) = \tfrac{\sqrt3}{2}

for

0 \le \theta < 2\pi

1Let

x = \theta + \tfrac{\pi}{6}

, so

\tfrac{\pi}{6} \le x < \tfrac{13\pi}{6}

2Solutions of

\sin x = \tfrac{\sqrt3}{2}

x = \tfrac{\pi}{3}

and

x = \tfrac{2\pi}{3}

; the next ones,

\tfrac{7\pi}{3}

and beyond, are too big.

3Subtract

\tfrac{\pi}{6}

Answer

\theta = \tfrac{\pi}{6}

\theta = \tfrac{\pi}{2}

Why this works

Changing the interval first is the whole trick. Students who solve for $x$ in the original interval and then divide systematically lose half the solutions of $\sin 2\theta = k$ .

Quadratics in a trigonometric function

Equations like $2\sin^{2}\theta - \sin\theta - 1 = 0$ are quadratics in $\sin\theta$ . Factorise (or use the formula) treating $\sin\theta$ as the unknown.

Each root gives a separate trigonometric equation. Reject any root outside the range of the function — for instance $\sin\theta = 2$ has no solutions.

Never divide both sides by $\sin\theta$ or $\cos\theta$ : you lose the solutions where that function is zero. Factorise instead.

Example 1Solve

2\sin^{2}\theta - \sin\theta - 1 = 0

for

0^{\circ} \le \theta < 360^{\circ}

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

\sin\theta = 1

gives

\theta = 90^{\circ}

\sin\theta = -\tfrac12

: principal value

-30^{\circ}

; in the interval the solutions are

180^{\circ} + 30^{\circ}

and

360^{\circ} - 30^{\circ}

Answer

\theta = 90^{\circ}, 210^{\circ}, 330^{\circ}

Using identities first

When an equation mixes functions, use an identity to rewrite it in terms of one. The usual choices: replace $\cos^{2}\theta$ with $1 - \sin^{2}\theta$ (or vice versa); replace $\dfrac{\sin\theta}{\cos\theta}$ with $\tan\theta$ ; replace $\sec^{2}\theta$ with $1 + \tan^{2}\theta$ ; expand $\sin 2\theta$ or $\cos 2\theta$ .

Equations like $a\sin\theta + b\cos\theta = c$ are solved by first writing the left side in harmonic form $R\sin(\theta + \alpha)$ .

Example 1Solve

3\cos^{2}\theta = 2 + 3\sin\theta

for

0^{\circ} \le \theta < 360^{\circ}

1Replace

\cos^{2}\theta

3(1 - \sin^{2}\theta) = 2 + 3\sin\theta

2Rearrange:

3\sin^{2}\theta + 3\sin\theta - 1 = 0

\sin\theta = \dfrac{-3 \pm \sqrt{9 + 12}}{6}

, giving

0.2638

-1.2638

(rejected, outside

[-1, 1]

\sin\theta = 0.2638

\theta = 15.30^{\circ}

180^{\circ} - 15.30^{\circ}

Answer

\theta = 15.3^{\circ}

164.7^{\circ}

Tip — If $\sin 2\theta$ appears alongside $\sin\theta$ or $\cos\theta$ , expand it as $2\sin\theta\cos\theta$ and factorise — do not cancel.

Think like an examiner

Show the adjusted interval whenever the argument is not simply $\theta$ .
List every solution, then check each lies in the required interval.
Give answers in the unit the interval uses — degrees or radians.
Explain rejection of impossible roots such as $\sin\theta = -1.26$ .

Common misconceptions

Stopping at the calculator value.

Use symmetry and the period to find every solution in the interval.

Dividing

\sin\theta\cos\theta = \sin\theta

\sin\theta

Factorise:

\sin\theta(\cos\theta - 1) = 0

, keeping

\sin\theta = 0

Solving

\sin 2\theta = k

over

0^{\circ}

360^{\circ}

for

2\theta

The interval for

2\theta

0^{\circ}

720^{\circ}

Using the cosine symmetry

180^{\circ} - \alpha

For cosine the second solution is

360^{\circ} - \alpha

Trigonometric equations

\sin\theta = k \Rightarrow \alpha,\ 180^{\circ} - \alpha

Sine pair.

\cos\theta = k \Rightarrow \alpha,\ 360^{\circ} - \alpha

Cosine pair.

\tan\theta = k \Rightarrow \alpha + 180^{\circ}n

Tangent repeats.

Stretch yourself

Solve $\sin 2\theta = \cos\theta$ for $0 \le \theta \le 2\pi$ , giving exact answers.

Hint — Expand $\sin 2\theta$ , move everything to one side and factorise out $\cos\theta$ .

Reveal solution

Solution

2\sin\theta\cos\theta - \cos\theta = 0

, so

\cos\theta(2\sin\theta - 1) = 0

. Either

\cos\theta = 0

, giving

\theta = \tfrac{\pi}{2}

\tfrac{3\pi}{2}

; or

\sin\theta = \tfrac12

, giving

\theta = \tfrac{\pi}{6}

\tfrac{5\pi}{6}

. So

\theta = \tfrac{\pi}{6}, \tfrac{\pi}{2}, \tfrac{5\pi}{6}, \tfrac{3\pi}{2}

. Dividing by

\cos\theta

at the start would have lost

\tfrac{\pi}{2}

and

\tfrac{3\pi}{2}

. Check

\theta = \tfrac{5\pi}{6}

\sin\tfrac{5\pi}{3} = -\tfrac{\sqrt3}{2}

and

\cos\tfrac{5\pi}{6} = -\tfrac{\sqrt3}{2}

✓.

\blacksquare

Questions students ask

How do I know how many solutions to expect?

Sketch the graph over the interval and draw the line

y = k

. Each intersection is one solution.

Why can’t I divide by

\cos\theta

\cos\theta = 0

is a solution, dividing by it removes that solution. Factorising keeps it.

Degrees or radians?

Match the interval in the question. If it involves

\pi

, work and answer in radians.

What if the principal value is negative?

Add a full period to bring it into the interval, and use symmetry to find the partner solution.

Key takeaways

Find the principal value, then the symmetric partner, then add periods.
For multiple or shifted angles, transform the interval before solving.
Quadratics in a trig function: factorise, then solve each factor, rejecting impossible values.
Use identities to reduce to a single function; never divide by a trig function that could be zero.
Answer in the units of the interval.

How this fits the course

Build on