Solving Trigonometric Equations — Multiple Angles
$\sin\theta = \tfrac{1}{2}$ has 2 solutions in $[0, 2\pi)$. But $\sin(2\theta) = \tfrac{1}{2}$ has 4, and $\sin(3\theta) = \tfrac{1}{2}$ has 6. More cycles fit into the same interval — so you need more solutions. The trick is deceptively simple: expand the interval before you solve, then divide back at the end.
How many solutions does $\sin(2\theta) = \tfrac{1}{2}$ have in $0 \le \theta < 2\pi$? Is it the same number as $\sin\theta = \tfrac{1}{2}$? Write your gut answer before reading on — even if you're not sure.
Multiple-angle equations seem harder but follow a rigid four-step process. Master the steps and you'll never miss a solution:
- Substitute — let $u = n\theta$ (the multiple angle), where $n$ is the multiplier.
- Expand the interval — if the original interval is $[0, 2\pi)$, the interval for $u$ is $[0, 2n\pi)$. Multiply both bounds by $n$.
- Solve — find all values of $u$ in the expanded interval satisfying $\sin u = k$, $\cos u = k$, or $\tan u = k$.
- Divide — compute $\theta = u/n$ to get all solutions. Check each falls in the original interval.
Key facts
- For $\sin(n\theta) = k$ in $[0, 2\pi)$, solve for $n\theta$ in $[0, 2n\pi)$ then divide by $n$
- Both bounds of the interval are multiplied by $n$ — including a non-zero lower bound
- Multiple-angle equations generally have $n$ times as many solutions as single-angle equations
Concepts
- Why the trig function completes $n$ cycles in the original interval when the angle is $n\theta$
- How to count the expected number of solutions as a check on your working
- Why the substitution $u = n\theta$ is structurally identical to the method for single-angle equations
Skills
- Solve $\sin(n\theta) = k$, $\cos(n\theta) = k$, and $\tan(n\theta) = k$ in any specified interval
- Handle non-standard intervals (e.g. $-\pi < \theta \le \pi$, $0 \le \theta \le \pi$) by expanding both bounds
- Verify solutions by substituting back into the original equation
Let's work through the example from the hero section — fully and carefully — to establish the method.
Worked Example 1 — cosine, double angle: the key definition, formula, and conditions — write this into your book.
Pause — copy the four-step double-angle equation method into your book as shown in Worked Example 1: choose the $\cos 2\theta$ form that matches remaining terms → substitute → solve quadratic → apply domain check.
Solve $\cos(2\theta) = \dfrac{1}{2}$ for $0 \le \theta < 2\pi$.
Interval for $u$: since $0 \le \theta < 2\pi$, multiply by 2: $0 \le u < 4\pi$.
Reference angle: $\dfrac{\pi}{3}$. $\cos > 0$ in Q1 and Q4 (per cycle).
Cycle 1 ($[0, 2\pi)$): $u = \dfrac{\pi}{3},\; \dfrac{5\pi}{3}$
Cycle 2 ($[2\pi, 4\pi)$): $u = 2\pi + \dfrac{\pi}{3} = \dfrac{7\pi}{3},\; 2\pi + \dfrac{5\pi}{3} = \dfrac{11\pi}{3}$
$\theta = \dfrac{\pi}{6},\; \dfrac{5\pi}{6},\; \dfrac{7\pi}{6},\; \dfrac{11\pi}{6}$
Quick check: When solving $\sin(2\theta) = \tfrac{\sqrt{3}}{2}$ for $0 \le \theta < 2\pi$, what is the correct interval to use for $u = 2\theta$?
Solve $\tan(3\theta) = 1$ for $0 \le \theta < \pi$.
Interval for $u$: $0 \le \theta < \pi$ multiplied by 3 gives $0 \le u < 3\pi$.
$\tan u = 1$ has period $\pi$, so one solution per $\pi$-interval.
$u = \dfrac{\pi}{4},\; \dfrac{\pi}{4} + \pi = \dfrac{5\pi}{4},\; \dfrac{\pi}{4} + 2\pi = \dfrac{9\pi}{4}$
$\theta = \dfrac{\pi}{12},\; \dfrac{5\pi}{12},\; \dfrac{9\pi}{12} = \dfrac{3\pi}{4}$
Did you get this? True or false: $\sin(2\theta) = \tfrac{1}{2}$ has the same solutions as $\sin\theta = \tfrac{1}{2}$ in $[0, 2\pi)$.
Worked examples · 3 in a row, reveal as you go
Solve $\cos(2\theta) = -\dfrac{1}{2}$ for $-\pi < \theta \le \pi$.
Expand: since $-\pi < \theta \le \pi$, multiply by 2: $-2\pi < u \le 2\pi$.
Reference angle: $\dfrac{\pi}{3}$. $\cos < 0$ in Q2 and Q3.
In $[0, 2\pi]$: $u = \dfrac{2\pi}{3},\; \dfrac{4\pi}{3}$
In $(-2\pi, 0)$: $u = -\dfrac{2\pi}{3},\; -\dfrac{4\pi}{3}$
$\theta = \dfrac{\pi}{3},\; \dfrac{2\pi}{3},\; -\dfrac{\pi}{3},\; -\dfrac{2\pi}{3}$
Fill the gap: To solve $\sin(2\theta) = \tfrac{\sqrt{3}}{2}$ for $0 \le \theta < 2\pi$, the correct interval for $u = 2\theta$ is $0 \le u < $ .
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: to solve $\tan(3\theta) = \sqrt{3}$ for $0 \le \theta < \pi$, the expanded interval for $u = 3\theta$ is $0 \le u < 3\pi$.
Activities · practice the four steps
Solve $\sin(2\theta) = \dfrac{\sqrt{3}}{2}$ for $0 \le \theta < 2\pi$. How many solutions should you expect?
Solve $\tan(3\theta) = 1$ for $0 \le \theta < \pi$. Show the expanded interval clearly.
Solve $\cos(2\theta) = -\dfrac{1}{2}$ for $-\pi < \theta \le \pi$. What is the expanded interval?
A student claims $\cos(2\theta) = \tfrac{1}{2}$ has only 2 solutions in $[0, 2\pi)$. What mistake did they make? Correct their work.
Explain in your own words why $\sin(n\theta) = k$ has $n$ times as many solutions as $\sin\theta = k$ in $[0, 2\pi)$.
Odd one out: Three of these statements about solving $\sin(2\theta) = \tfrac{1}{2}$ in $[0, 2\pi)$ are correct. Which is WRONG?
At the start you were asked how many solutions $\sin(2\theta) = \tfrac{1}{2}$ has in $[0, 2\pi)$. The answer is 4 — not 2.
Letting $u = 2\theta$ and working in $[0, 4\pi)$: $\sin u = \tfrac{1}{2}$ gives reference angle $\tfrac{\pi}{6}$, and $\sin > 0$ in Q1 and Q2. In cycle 1: $u = \tfrac{\pi}{6}, \tfrac{5\pi}{6}$. In cycle 2: $u = \tfrac{13\pi}{6}, \tfrac{17\pi}{6}$. Dividing by 2: $\theta = \tfrac{\pi}{12}, \tfrac{5\pi}{12}, \tfrac{13\pi}{12}, \tfrac{17\pi}{12}$.
Why does replacing $\theta$ with $n\theta$ give $n$ times as many solutions in the same interval?
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Solve $\sin(2\theta) = \dfrac{\sqrt{3}}{2}$ for $0 \le \theta < 2\pi$. (3 marks)
Q2. Solve $\tan(3\theta) = 1$ for $0 \le \theta < \pi$. (3 marks)
Q3. Solve $\cos(2\theta) = -\dfrac{1}{2}$ for $-\pi < \theta \le \pi$. (3 marks)
Comprehensive answers (click to reveal)
Activity 1:
1. $u = 2\theta \in [0,4\pi)$. $\sin u = \frac{\sqrt{3}}{2}$, ref angle $\frac{\pi}{3}$. Cycle 1: $u = \frac{\pi}{3}, \frac{2\pi}{3}$. Cycle 2: $u = \frac{\pi}{3}+2\pi = \frac{7\pi}{3}$, $u = \frac{2\pi}{3}+2\pi = \frac{8\pi}{3}$. $\theta = \frac{\pi}{6}, \frac{\pi}{3}, \frac{7\pi}{6}, \frac{4\pi}{3}$.
2. $u = 3\theta \in [0,3\pi)$. $\tan u = 1$, ref angle $\frac{\pi}{4}$. $u = \frac{\pi}{4}, \frac{5\pi}{4}, \frac{9\pi}{4}$. $\theta = \frac{\pi}{12}, \frac{5\pi}{12}, \frac{3\pi}{4}$.
3. $u = 2\theta \in (-2\pi,2\pi]$. $\cos u = -\frac{1}{2}$, ref angle $\frac{\pi}{3}$. $u = \frac{2\pi}{3}, \frac{4\pi}{3}, -\frac{2\pi}{3}, -\frac{4\pi}{3}$. $\theta = \frac{\pi}{3}, \frac{2\pi}{3}, -\frac{\pi}{3}, -\frac{2\pi}{3}$.
4. Mistake: only used $[0, 2\pi)$ for $u = 2\theta$ instead of $[0, 4\pi)$. Correct: $\theta = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$.
Q1 (3 marks): Correct expanded interval $[0,4\pi)$ stated [1]. All four values of $u$ found [1]. Correct $\theta$ values: $\frac{\pi}{6}, \frac{\pi}{3}, \frac{7\pi}{6}, \frac{4\pi}{3}$ [1].
Q2 (3 marks): Correct expanded interval $[0,3\pi)$ [1]. Three $u$ values identified [1]. Correct $\theta$ values: $\frac{\pi}{12}, \frac{5\pi}{12}, \frac{3\pi}{4}$ [1].
Q3 (3 marks): Correct expanded interval $(-2\pi, 2\pi]$ [1]. All four $u$ values (positive and negative) found [1]. Correct $\theta$ values: $\pm\frac{\pi}{3}, \pm\frac{2\pi}{3}$ [1].
Five timed questions on multiple-angle trig equations. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering multiple-angle trig questions. Lighter alternative to the boss.
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