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hscscience Maths Ext 1 · Y11
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Module 3 · L10 of 15 ~35 min ⚡ +95 XP available

Solving Trigonometric Equations, Linear

Every trig equation of the form $\sin\theta = k$, $\cos\theta = k$, or $\tan\theta = k$ has infinitely many solutions spread across the number line, but within any finite interval you need a systematic approach. The ASTC rule and the reference angle technique give you a two-step process that never misses a solution and never produces a spurious one.

Today's hook, How many solutions does $\sin\theta = \tfrac{1}{2}$ have in $0 \le \theta < 2\pi$? Many students say "one", the calculator gives $\pi/6$ and that seems to be it. But there's a second solution hiding in a different quadrant. By the end of this lesson you'll find all of them instantly, every time.
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Recall, your gut answer first

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Recall, your gut answer first
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How many solutions does $\sin\theta = \tfrac{1}{2}$ have in $0 \le \theta < 2\pi$? Without using a calculator guess the number and name the solutions if you can. Then: how many solutions in $0 \le \theta < 4\pi$?

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Orientation and key terms

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The two key ideas
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Every trig equation reduces to two steps. Lock these in and you'll never miss a solution:

Step 1: find the reference angle the acute angle whose trig value (ignoring sign) equals $|k|$. Step 2: use ASTC to find which quadrants have solutions for the given sign of $k$.

Two-step trig solving flow: reference angle then ASTC quadrants
Ref angle → ASTC → write solutions in interval
ASTC rule
A ll positive in Q1, S ine positive in Q2, T an positive in Q3, C os positive in Q4. Mnemonic: "All Stations To Central."
Sine/Cosine: 2 solutions
For $|k| < 1$, each of $\sin\theta = k$ and $\cos\theta = k$ has exactly 2 solutions per $2\pi$ period.
Tangent: 1 solution
$\tan\theta = k$ has exactly 1 solution per $\pi$ period, tangent is one-to-one over each half-period.
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What you'll master
Know

Key facts

  • General solution of $\sin\theta = k$: $\theta = n\pi + (-1)^n\alpha$
  • General solution of $\cos\theta = k$: $\theta = 2n\pi \pm \alpha$
  • General solution of $\tan\theta = k$: $\theta = n\pi + \alpha$
Understand

Concepts

  • Why sine and cosine have two solutions per period but tangent has one
  • The symmetry of the unit circle that generates the second solution
  • The difference between a principal value and a general solution
Can do

Skills

  • Find the reference angle from the principal value
  • Use ASTC to identify all quadrants containing solutions
  • List all solutions within a specified interval, including wider intervals spanning multiple periods
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Key terms
Reference angle $\alpha$The acute angle (between $0$ and $\pi/2$) whose trig value equals $|k|$. All solutions are expressed in terms of $\alpha$.
Principal valueThe value returned by the inverse trig function on a calculator; lies in the restricted range of that function.
General solutionA formula (using integer $n$) that generates all possible solutions across all periods.
ASTC ruleMnemonic for which trig functions are positive in each quadrant: All, Sine, Tan, Cos (Q1–Q4).
PeriodThe interval over which the function repeats: $2\pi$ for sine and cosine, $\pi$ for tangent.
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General solutions and the unit circle

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General solutions and the unit circle
core concept

The unit circle has symmetry that creates multiple solutions for the same trig value. For a given value $k$ with reference angle $\alpha$:

$$\sin\theta = k \;\Rightarrow\; \theta = n\pi + (-1)^n\alpha, \quad n \in \mathbb{Z}$$
$$\cos\theta = k \;\Rightarrow\; \theta = 2n\pi \pm \alpha, \quad n \in \mathbb{Z}$$
$$\tan\theta = k \;\Rightarrow\; \theta = n\pi + \alpha, \quad n \in \mathbb{Z}$$

In practice, for an exam, you rarely use the general solution directly. Instead, you:

  1. Find the reference angle $\alpha = \sin^{-1}|k|$, $\cos^{-1}|k|$, or $\tan^{-1}|k|$.
  2. Use ASTC to determine which quadrants apply.
  3. Write all solutions in the required interval by adding $2\pi$ (or $\pi$ for tangent) repeatedly.
Why two vs one? The sine curve reaches height $k$ on the way up (Q1 or Q2) and on the way down (also Q1 or Q2 by symmetry). Tangent is strictly increasing over each $(-\pi/2, \pi/2)$ interval, it can only hit each value exactly once per period.

Reference angle first: = ^{-1}|k| (or ^{-1}|k| or ^{-1}|k|); ASTC: All positive Q1, Sine Q2, Tan Q3, Cos Q4

Pause, copy the general solution method into your book: find the reference angle $\alpha = \arcsin|k|$ (or $\arccos|k|$ or $\arctan|k|$), then use ASTC to list all quadrants with the correct sign, giving two solutions per period.

Quick check: How many solutions does $\cos\theta = \tfrac{1}{2}$ have in $0 \le \theta < 2\pi$?

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Worked examples

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · COSINE EQUATION

Solve $\cos\theta = \dfrac{\sqrt{3}}{2}$ for $0 \le \theta < 2\pi$.

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Reference angle: $\alpha = \cos^{-1}\!\left(\dfrac{\sqrt{3}}{2}\right) = \dfrac{\pi}{6}$
The reference angle is the acute angle with $\cos\alpha = \sqrt{3}/2$. This is a known exact value.
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PROBLEM 2 · TANGENT EQUATION

Solve $\tan\theta = -1$ for $-\pi < \theta \le \pi$.

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Reference angle: $\alpha = \tan^{-1}(1) = \dfrac{\pi}{4}$
The reference angle uses $|\!-\!1| = 1$. $\tan(\pi/4) = 1$, so $\alpha = \pi/4$.
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PROBLEM 3 · WIDER INTERVAL

Solve $\sin\theta = \dfrac{\sqrt{3}}{2}$ for $0 \le \theta \le 4\pi$.

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Reference angle: $\alpha = \sin^{-1}\!\left(\dfrac{\sqrt{3}}{2}\right) = \dfrac{\pi}{3}$
$\sin(\pi/3) = \sqrt{3}/2$, an exact value to recognise.
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Did you get this? True or false: $\sin\theta = k$ always has exactly two solutions in $[0, 2\pi)$ for any $k$ with $|k| \le 1$.

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Misconceptions to fix

Misconceptions to fix · the 3 traps that cost marks

Trap 01
Only one solution from the calculator
The inverse trig button gives the principal value, one answer. But $\sin\theta = k$ has two solutions per period and $\cos\theta = k$ has two. Always use ASTC to check for the second solution.
Trap 02
Wrong symmetry formula for the second solution
For sine, the second solution is $\pi - \alpha$ (not $2\pi - \alpha$). For cosine it is $2\pi - \alpha$. Mixing these up is extremely common. Use ASTC to determine the quadrant first, then apply the correct formula.
Trap 03
Missing solutions in wider intervals
In an interval like $[0, 4\pi]$, there are four sine or cosine solutions. After finding the two in $[0, 2\pi)$, add $2\pi$ to each. Always state how many solutions you expect and check you have that many.

Fill the gap: For $\sin\theta = k$ in $[0, 2\pi)$, if the reference angle is $\alpha$, the two solutions are $\alpha$ and .

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Activities

Activities · practice with the ideas

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Solve $\sin\theta = \dfrac{\sqrt{3}}{2}$ for $0 \le \theta < 2\pi$.

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Solve $\cos\theta = -\dfrac{1}{2}$ for $0 \le \theta < 2\pi$.

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Solve $\tan\theta = \sqrt{3}$ for $-\dfrac{\pi}{2} < \theta < \dfrac{\pi}{2}$.

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Solve $\sin\theta = -\dfrac{1}{\sqrt{2}}$ for $0 \le \theta < 2\pi$. State the quadrants and give exact answers.

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Explain why $\tan\theta = k$ has exactly one solution per $\pi$ interval, while $\sin\theta = k$ has two per $2\pi$ interval.

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Revisit your thinking

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Revisit your thinking

Earlier you were asked how many solutions $\sin\theta = \tfrac{1}{2}$ has in $[0, 2\pi)$.

The answer is two: $\theta = \dfrac{\pi}{6}$ (Q1) and $\theta = \pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6}$ (Q2). In $[0, 4\pi)$ there are four: $\dfrac{\pi}{6},\, \dfrac{5\pi}{6},\, \dfrac{\pi}{6}+2\pi = \dfrac{13\pi}{6},\, \dfrac{5\pi}{6}+2\pi = \dfrac{17\pi}{6}$.

The key insight: sine is symmetric about $\theta = \pi/2$, so both $\pi/6$ and $5\pi/6$ give the same sine value. Tangent has no such symmetry within a half-period, hence one solution per $\pi$.

Check: True or false: $\sin\theta = -1$ has exactly one solution in $[0, 2\pi)$.

Odd one out: Which statement is the odd one out (i.e. incorrect)?

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Multiple choice

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Multiple choice
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Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.

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Short answer

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Short answer
ApplyBand 42 marks

Q1. Solve $\sin\theta = \dfrac{\sqrt{3}}{2}$ for $0 \le \theta < 2\pi$, giving exact answers. (2 marks)

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ApplyBand 42 marks

Q2. Solve $\cos\theta = -\dfrac{1}{2}$ for $0 \le \theta < 2\pi$, giving exact answers. (2 marks)

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ApplyBand 31 mark

Q3. Solve $\tan\theta = \sqrt{3}$ for $-\dfrac{\pi}{2} < \theta < \dfrac{\pi}{2}$. (1 mark)

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Comprehensive answers (click to reveal)

Activity: 1. $\alpha = \pi/3$; Q1 and Q2; $\theta = \pi/3, 2\pi/3$ · 2. $\alpha = \pi/3$; Q2 and Q3; $\theta = 2\pi/3, 4\pi/3$ · 3. $\alpha = \pi/3$; tan positive in Q1; only $\theta = \pi/3$ in $(-\pi/2, \pi/2)$ · 4. $\alpha = \pi/4$; sine negative in Q3 and Q4; $\theta = \pi + \pi/4 = 5\pi/4$ and $\theta = 2\pi - \pi/4 = 7\pi/4$ · 5. Tangent is strictly increasing over each interval $(-\pi/2, \pi/2)$ so it hits each value exactly once; sine reaches the same height twice per $2\pi$ period (once ascending, once descending).

Q1 (2 marks): $\alpha = \pi/3$ [0.5]. Sine positive in Q1 and Q2. Q1: $\theta = \pi/3$ [0.5]. Q2: $\theta = \pi - \pi/3 = 2\pi/3$ [1].

Q2 (2 marks): $\alpha = \pi/3$ [0.5]. Cosine negative in Q2 and Q3. Q2: $\theta = \pi - \pi/3 = 2\pi/3$ [0.5]; Q3: $\theta = \pi + \pi/3 = 4\pi/3$ [1].

Q3 (1 mark): $\alpha = \pi/3$; $\tan\theta > 0$ in Q1; only Q1 solution in $(-\pi/2, \pi/2)$: $\theta = \pi/3$ [1].

Module overview · Extension 1 · Checkpoint 3 · Module quiz