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.
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?
Every quadratic-type trig equation follows the same five steps. Memorise the flow and you can handle any variation:
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.
Solve $2\sin^2\theta - \sin\theta - 1 = 0$ for $0 \le \theta < 2\pi$.
Quick check: When solving a quadratic trig equation and one root gives $\cos\theta = 1.5$, what should you do?
Solve $\sec^2\theta - 3\sec\theta + 2 = 0$ for $0 \le \theta < 2\pi$.
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
Solve $\sin^2\theta = 3\cos^2\theta$ for $0 \le \theta < 2\pi$.
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
Did you get this? True or false: the equation $2\sin^2\theta - \sin\theta - 1 = 0$ has exactly 2 solutions in $[0, 2\pi)$.
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?
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?
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 $2\cos^2\theta + \cos\theta - 1 = 0$ for $0 \le \theta < 2\pi$. (3 marks)
Q2. Solve $\tan^2\theta - \tan\theta - 2 = 0$ for $0 \le \theta < \pi$. (3 marks)
Q3. Solve $\sin^2\theta = 3\cos^2\theta$ for $0 \le \theta < 2\pi$. (3 marks)
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].
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