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Module 3 · L11 of 15 ~35 min ⚡ +95 XP available

Solving Trigonometric Equations — Quadratic Type

You already know how to solve $\sin\theta = \frac{1}{2}$ — but what about $2\sin^2\theta - \sin\theta - 1 = 0$? This is a quadratic disguised as a trig equation. The trick is a single substitution that turns a trig puzzle into a polynomial you already know how to solve — then you use the unit circle to finish the job.

Today's hook — Staring at $2\sin^2\theta - \sin\theta - 1 = 0$ in an exam can feel paralysing. But the moment you write $u = \sin\theta$ it becomes $2u^2 - u - 1 = 0$ — a plain quadratic you can factorise in seconds. One substitution is all it takes. By the end of this lesson you'll do it automatically.

0/5QUESTS

Recall — your gut answer first

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How would you solve $2\sin^2\theta - \sin\theta - 1 = 0$? Without reading ahead — what substitution might help, and what do you think the solutions look like?

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The five-step method

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Every quadratic-type trig equation follows the same five steps. Memorise the flow and you can handle any variation:

Identify — the equation is quadratic in a single trig function ($\sin\theta$, $\cos\theta$, $\tan\theta$, $\sec\theta$, etc.).
Substitute — let $u =$ that trig function to form a polynomial equation in $u$.
Solve — factorise or use the quadratic formula to find all values of $u$.
Reject — discard any $u$ outside the range of that trig function ($[-1,1]$ for $\sin$ and $\cos$; all real for $\tan$).
Solve again — for each valid $u$, solve the resulting linear trig equation in the given interval.

$2\sin^2\theta - \sin\theta - 1 = 0$

Range of $\sin$ and $\cos$

Both are bounded: $-1 \le \sin\theta \le 1$ and $-1 \le \cos\theta \le 1$. Any $u$ value outside this range has no solutions — reject it immediately.

$\tan$ has no range restriction

$\tan\theta$ can take any real value, so if the quadratic is in $\tan\theta$, all values of $u$ from the quadratic are potentially valid — no rejection step needed for range.

Factorise or use the formula

Most HSC quadratic trig equations have integer or simple rational roots — try factorisation first. Use the quadratic formula only if factorisation fails.

What you'll master

Know

Key facts

Quadratic-type trig equations require a substitution to convert to polynomial form
$\sin\theta$ and $\cos\theta$ are bounded: range $[-1, 1]$; values outside are rejected
The five-step method: identify, substitute, solve, reject, solve trig

Understand

Concepts

Why solutions from the polynomial step may not correspond to valid angles
How the unit circle is used after substitution to find all angles in the interval
When to use $\sin^2\theta + \cos^2\theta = 1$ to convert to a single trig function

Can do

Skills

Solve quadratic trig equations using substitution within specified intervals
Recognise and handle equations involving $\sec$, $\csc$, $\cot$ as the quadratic variable
Use $\sin^2\theta = 1 - \cos^2\theta$ to reduce mixed equations to a single function

Key terms

Quadratic typeAn equation that is quadratic in form when a substitution replaces the trig function with a variable $u$.

SubstitutionReplacing a trig function with a single variable, e.g. let $u = \sin\theta$, to create a polynomial equation.

RejectionDiscarding values of $u$ that lie outside the range of the trig function — these correspond to no real angle.

Pythagorean identity$\sin^2\theta + \cos^2\theta = 1$, used to convert equations with both $\sin^2\theta$ and $\cos^2\theta$ to a single function.

Worked Example 1 — basic substitution with $\sin$

core concept

The clearest way to learn the method is to work through it step by step. Here is the equation from the Think First question, fully solved.

Worked Example 1 — basic substitution with $\sin$: the key definition, formula, and conditions — write this into your book.

Pause — copy the five-step trig equation method into your book as shown in Worked Example 1: rearrange → identify substitution → write the domain → list solutions for $u$ → back-substitute.

PROBLEM 1 · QUADRATIC IN $\sin\theta$

Solve $2\sin^2\theta - \sin\theta - 1 = 0$ for $0 \le \theta < 2\pi$.

Let $u = \sin\theta$. The equation becomes $2u^2 - u - 1 = 0$.

The equation is quadratic in $\sin\theta$. Substitution converts it to a polynomial.

Quick check: When solving a quadratic trig equation and one root gives $\cos\theta = 1.5$, what should you do?

PROBLEM 2 · QUADRATIC IN $\sec\theta$

Solve $\sec^2\theta - 3\sec\theta + 2 = 0$ for $0 \le \theta < 2\pi$.

Let $u = \sec\theta$: $u^2 - 3u + 2 = 0$

The equation is quadratic in $\sec\theta$.

Did you get this? True or false: when solving a quadratic in $\sec\theta$ and one solution gives $\sec\theta = 0.5$, this root should be rejected.

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

PROBLEM 3 · REDUCING USING $\sin^2\theta + \cos^2\theta = 1$

Solve $\sin^2\theta = 3\cos^2\theta$ for $0 \le \theta < 2\pi$.

$\dfrac{\sin^2\theta}{\cos^2\theta} = 3 \;\Rightarrow\; \tan^2\theta = 3$

Divide both sides by $\cos^2\theta$ (valid since we can check $\cos\theta \ne 0$ separately). This converts to a single trig function.

Fill the gap: To solve $2\cos^2\theta + \cos\theta - 1 = 0$, let $u = \cos\theta$ to get $2u^2 + u - 1 = 0$. This factorises as $(2u - 1)(u + 1) = 0$, giving $u = \tfrac{1}{2}$ or $u = $ .

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Forgetting to reject out-of-range values

After factorising, a student writes down all polynomial roots and tries to solve the trig equation for each one — including $\sin\theta = 2$ or $\cos\theta = -3$. These have no solutions. Always check $-1 \le u \le 1$ for $\sin$ and $\cos$ before proceeding. Skipping the rejection step will cost you marks for writing down invalid angle solutions.

Trap 02

Losing solutions by taking only the positive square root

$\sin^2\theta = \tfrac{1}{4}$ gives $\sin\theta = \pm\tfrac{1}{2}$ — not just $+\tfrac{1}{2}$. Students who write only $\sin\theta = \tfrac{1}{2}$ will find 2 solutions instead of 4 in $[0, 2\pi)$. Similarly, $\tan^2\theta = 3$ gives $\tan\theta = \pm\sqrt{3}$, doubling the number of solutions.

Trap 03

Mixing $\sin^2\theta$ and $\cos^2\theta$ without reducing

Substitution only works if the equation is in a single trig function. If you have both $\sin^2\theta$ and $\cos^2\theta$ in the same equation, use $\sin^2\theta + \cos^2\theta = 1$ to eliminate one function first before substituting. Trying to substitute both simultaneously is not valid.

Did you get this? True or false: the equation $2\sin^2\theta - \sin\theta - 1 = 0$ has exactly 2 solutions in $[0, 2\pi)$.

Work mode · how are you completing this lesson?

Activities · practice with the method

Solve $2\cos^2\theta + \cos\theta - 1 = 0$ for $0 \le \theta < 2\pi$. Show all five steps.

Solve $\tan^2\theta - \tan\theta - 2 = 0$ for $0 \le \theta < \pi$. How many solutions are there?

Solve $\sin^2\theta = 3\cos^2\theta$ for $0 \le \theta < 2\pi$. (Hint: divide both sides by $\cos^2\theta$.)

A student solves $2\sin^2\theta + 3\sin\theta - 2 = 0$ and gets $\sin\theta = \frac{1}{2}$ and $\sin\theta = -2$. What should they do with the second root? What are the final solutions in $[0, 2\pi)$?

Explain in your own words why a quadratic trig equation can have fewer solutions than expected. Give an example.

Odd one out: Three of these are valid steps in solving a quadratic trig equation. Which is NOT?

Revisit your thinking

At the start you were asked how you would tackle $2\sin^2\theta - \sin\theta - 1 = 0$. You've now seen that the substitution $u = \sin\theta$ is the key — turning a potentially unfamiliar trig equation into a straightforward quadratic.

The solutions were $\theta = \dfrac{\pi}{2}, \dfrac{7\pi}{6}, \dfrac{11\pi}{6}$, because $\sin\theta = 1$ gives one angle and $\sin\theta = -\tfrac{1}{2}$ gives two angles in the third and fourth quadrants. Why must we check whether each value of $u$ is in $[-1,1]$ before finding the angles?

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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.

Short answer

ApplyBand 43 marks

Q1. Solve $2\cos^2\theta + \cos\theta - 1 = 0$ for $0 \le \theta < 2\pi$. (3 marks)

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ApplyBand 43 marks

Q2. Solve $\tan^2\theta - \tan\theta - 2 = 0$ for $0 \le \theta < \pi$. (3 marks)

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AnalyseBand 53 marks

Q3. Solve $\sin^2\theta = 3\cos^2\theta$ for $0 \le \theta < 2\pi$. (3 marks)

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Comprehensive answers (click to reveal)

Activity 1:

1. $2u^2+u-1=0 \Rightarrow (2u-1)(u+1)=0$; $u=\frac{1}{2}$ or $u=-1$. Both in $[-1,1]$. $\cos\theta=\frac{1}{2}$: $\theta=\frac{\pi}{3}, \frac{5\pi}{3}$. $\cos\theta=-1$: $\theta=\pi$. Solutions: $\theta=\frac{\pi}{3}, \pi, \frac{5\pi}{3}$.

2. $u^2-u-2=0\Rightarrow(u-2)(u+1)=0$; $u=2$ or $u=-1$. No range restriction for $\tan$. $\tan\theta=2$: $\theta=\tan^{-1}(2)\approx1.107$ rad (Q1) and $\pi+\tan^{-1}(2)\approx4.249$ rad (Q3). But interval is $[0,\pi)$, so only $\theta=\tan^{-1}(2)$. $\tan\theta=-1$: reference angle $\frac{\pi}{4}$; $\tan<0$ in Q2: $\theta=\frac{3\pi}{4}$. Solutions: $\theta=\tan^{-1}(2),\frac{3\pi}{4}$.

3. $\tan^2\theta=3$; $\tan\theta=\pm\sqrt{3}$. $\tan\theta=\sqrt{3}$: $\theta=\frac{\pi}{3}, \frac{4\pi}{3}$. $\tan\theta=-\sqrt{3}$: $\theta=\frac{2\pi}{3}, \frac{5\pi}{3}$. Solutions: $\theta=\frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}$.

4. Reject $\sin\theta=-2$ (outside $[-1,1]$). $\sin\theta=\frac{1}{2}$: $\theta=\frac{\pi}{6}, \frac{5\pi}{6}$.

Q1 (3 marks): Correct substitution and factorisation [1]. Both valid roots identified (neither rejected) [1]. All three solutions $\frac{\pi}{3}, \pi, \frac{5\pi}{3}$ [1].

Q2 (3 marks): Correct substitution and factorisation [1]. Both values of $u$ valid (no range restriction for $\tan$); selects angles in $[0,\pi)$ [1]. Solutions $\theta=\tan^{-1}2, \frac{3\pi}{4}$ [1].

Q3 (3 marks): Correct reduction to $\tan^2\theta=3$ [1]. Both $\pm\sqrt{3}$ considered [1]. All four solutions $\frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}$ [1].

Boss battle · The Trig Solver

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Five timed questions on quadratic trig equations. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

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Science Jump · platform challenge

Climb platforms by answering trig questions. Lighter alternative to the boss.

Mark lesson as complete

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Module overview · Extension 1 subject page · Checkpoint 3 · Module quiz