Trigonometric Equations with Multiple Angles

Video lesson · from 17:142 worked examples

One lesson video covers all of Trigonometry, so it opens at 17:14 for this topic — not from the beginning.

Theory

Multiple angles occur when the period of the given function is not $360^\circ$ or $2\pi$ .

The period is the length of one cycle of the given trigonometric function.

$y = \sin 2x$ has a period of $\pi$ radians.
$y = \cos 4x^\circ$ has a period of $90^\circ$ .

When solving multiple angle equations, adjust your domain to match the argument. E.g. if the domain is $0 \leq x \leq 360$ , solve $0 \leq 2x \leq 720$ first, then divide your answers by 2.

⚠️ Common Examiner Traps

Widen the domain before you solve: for $\sin 2x$ with $0 \leq x \leq 360$ , solve for $2x$ over $0 \leq 2x \leq 720$ . Forgetting this is the classic way to lose half the solutions.
Divide at the very end: find all the values of $2x$ first, then halve each one.
Count what you expect: $\sin 2x = k$ normally has four solutions in a full revolution, $\sin 3x = k$ six. If you have fewer, you have missed some.
Keep going round: add $360^\circ$ repeatedly to each base solution until you pass the widened domain.

Worked examples

Example 1

Solve $2\sin 2x^\circ - 1 = 0$ for $0 < x^\circ < 360$ .

First, isolate the trigonometric function:

\begin{aligned} 2\sin 2x^\circ &= 1 \\ \sin 2x^\circ &= \frac{1}{2} \end{aligned}

Since the domain is $0 < x < 360$ , the domain for $2x$ is $0 < 2x < 720$ . We're looking for solutions up to $720^\circ$ .

Base angle for $2x$ is $\sin^{-1}\left(\frac{1}{2}\right) = 30^\circ$ . Sine is positive in Q1, Q2.

\begin{aligned} 2x &= 30, 150 \quad \text{(First cycle)} \\ 2x &= 30 + 360 \\ &= 390 \quad \text{(Second cycle)} \\ 2x &= 150 + 360 \\ &= 510 \quad \text{(Second cycle)} \end{aligned}

Now, divide all answers by 2 to find $x$ :

x = 15^\circ, 75^\circ, 195^\circ, 255^\circ

Example 2

Solve $\sqrt{2}\cos 2x = 1$ for $0 < x < \pi$ .

\cos 2x = \frac{1}{\sqrt{2}}

Domain for $2x$ is $0 < 2x < 2\pi$ .

Base angle for $2x$ is $\cos^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}$ . Cosine is positive in Q1, Q4.

\begin{aligned} 2x &= \frac{\pi}{4}, \frac{7\pi}{4} \\ x &= \frac{\pi}{8}, \frac{7\pi}{8} \end{aligned}