If an equation contains a squared ratio, treat that ratio as a single letter. What is left is a quadratic you already know how to solve.
$2\sin^2\theta - \sin\theta - 1 = 0$ looks unfamiliar until you write $s$ for $\sin\theta$. Then it is $2s^2 - s - 1 = 0$, and you have been factorising that since Year 10.
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Recall, your gut answer first
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Recall, your gut answer first
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Look at $2\sin^2\theta - \sin\theta - 1 = 0$. Replace every $\sin\theta$ with the letter $s$. What kind of equation is left, and can you factorise it?
Before you work it out, what is your instinct? Write it down, then check it against the lesson.
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Substitute, solve, then unwind
Work through the core explanation before applying it.
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Substitute, solve, then unwind
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Let $s = \sin\theta$. Solve the resulting quadratic for $s$, discard any root outside $[-1, 1]$, then solve $\sin\theta = s$ on the domain for each root that survives.
$a\sin^2\theta + b\sin\theta + c = 0 \;\Rightarrow\; as^2 + bs + c = 0$, then solve $\sin\theta = s$
Reject roots outside $[-1, 1]$
Sine and cosine can never exceed $1$ or fall below $-1$. A root such as $s = 2$ is discarded with a reason, not carried forward. Tangent has no such restriction.
Use the identity to get one ratio
If an equation mixes $\sin^2\theta$ and $\cos\theta$, replace $\sin^2\theta$ with $1 - \cos^2\theta$ so everything is in cosine. A quadratic in two different ratios cannot be solved by substitution.
Each surviving root is a full equation
Two roots means two trigonometric equations, each with its own quadrant work and its own set of solutions. Stopping at the roots earns none of those marks.
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What you'll master
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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What you'll master
Know
Key facts
A trigonometric equation containing a squared ratio can become a quadratic by substitution.
Roots outside $[-1, 1]$ are rejected for sine and cosine.
The Pythagorean identity converts $\sin^2\theta$ into $1 - \cos^2\theta$ and back.
Each accepted root generates its own trigonometric equation to solve on the domain.
Understand
Concepts
Why substituting a single letter makes the structure visible.
Why a root outside $[-1, 1]$ has no solutions rather than being an error.
Why an equation mixing two different ratios must be converted before substituting.
Can do
Skills
Solve a trigonometric equation that is already a quadratic in one ratio.
Use the Pythagorean identity to reduce a mixed equation to one ratio.
Reject impossible roots with a reason and solve the rest on the domain.
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Key terms
Reducible to a quadraticAn equation that becomes a quadratic once one ratio is written as a single letter. Like this: $2\sin^2\theta - \sin\theta - 1 = 0$ becomes $2s^2 - s - 1 = 0$.
SubstitutionReplacing a repeated ratio with one letter. Like this: writing $c$ for $\cos\theta$ turns $2\cos^2\theta - 1 = 0$ into $2c^2 - 1 = 0$.
Rejecting a rootDiscarding a value the ratio cannot take, with a reason. Like this: $\sin\theta = 2$ has no solutions because sine never exceeds $1$.
Pythagorean identityThe relationship $\sin^2\theta + \cos^2\theta = 1$, used to swap one squared ratio for the other. Like this: $\sin^2\theta = 1 - \cos^2\theta$.
Unwinding the substitutionSolving the trigonometric equation for each root you kept. Like this: $s = -\dfrac{1}{2}$ gives $\sin\theta = -\dfrac{1}{2}$ to solve.
Restricted domainThe interval the answers must lie in. Like this: $0^\circ \leq \theta \leq 360^\circ$ keeps one revolution.
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Seeing the quadratic
Work through the core explanation before applying it.
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Seeing the quadratic
core concept
An equation such as $2\sin^2\theta - \sin\theta - 1 = 0$ contains only one ratio, $\sin\theta$, appearing squared and to the first power. Write $s = \sin\theta$ and it becomes $2s^2 - s - 1 = 0$.
That factorises as $(2s + 1)(s - 1) = 0$, so $s = -\dfrac{1}{2}$ or $s = 1$.
The substitution is a way of seeing, not a trick. Nothing changes about the equation; it just removes the trigonometric notation while you do the algebra.
$\sin^2\theta$ means $(\sin\theta)^2$. It is the ratio that is squared, not the angle. $\sin(\theta^2)$ is a different and much harder object, and it is not what this notation says.
Quick check: substituting $s = \sin\theta$ into $2\sin^2\theta - \sin\theta - 1 = 0$ gives which equation?
An equation containing one ratio, squared and to the first power, becomes a quadratic under the substitution $s = \sin\theta$. Solve for $s$ first. Note that $\sin^2\theta$ means $(\sin\theta)^2$.
Pause, copy the substitution, the factorisation $(2s + 1)(s - 1) = 0$, and the note on what $\sin^2\theta$ means, into your book.
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Rejecting the impossible root
core concept
We just saw how to get the roots. That raises a question: can every root be used? This card answers it → no, sine and cosine are trapped between $-1$ and $1$.
Sine and cosine of any angle lie in $[-1, 1]$. So a root such as $s = 2$ produces the equation $\sin\theta = 2$, which has no solutions at all.
Reject it explicitly and say why. "No solutions, since $\sin\theta$ cannot exceed $1$" is usually worth a mark; silently dropping it is not.
Tangent is different: it takes every real value, so a root of any size is acceptable when the substitution was $t = \tan\theta$.
Rejecting is part of the answer, not a step you skip. An equation whose quadratic has two roots, one of them impossible, still has a full solution set from the other. Show both roots and the reason one is discarded.
Which of these roots must be rejected?
Sine and cosine lie in $[-1, 1]$, so a root outside that range gives an equation with no solutions and is rejected with a stated reason. Tangent takes every real value, so no root is impossible for it.
Pause, copy the $[-1, 1]$ restriction, the wording used to reject a root, and the tangent exception, into your book.
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Mixed ratios, and the identity that fixes them
core concept
We just saw equations in one ratio. That raises a question: what if an equation contains both sine and cosine? This card answers it → use the Pythagorean identity to rewrite it in one of them.
An equation such as $2\sin^2\theta + 3\cos\theta = 3$ cannot be substituted directly, because it holds two different ratios.
Replace $\sin^2\theta$ with $1 - \cos^2\theta$: the equation becomes $2(1 - \cos^2\theta) + 3\cos\theta = 3$, then $-2\cos^2\theta + 3\cos\theta - 1 = 0$, or $2\cos^2\theta - 3\cos\theta + 1 = 0$.
Now the substitution works: $2c^2 - 3c + 1 = 0$ gives $(2c - 1)(c - 1) = 0$, so $c = \dfrac{1}{2}$ or $c = 1$, both of which are inside $[-1, 1]$.
Convert towards the ratio that appears unsquared. Here cosine appears to the first power, so rewrite the sine term. Converting the other way would leave a square root and no quadratic at all.
Fill the blank: using the Pythagorean identity, $\sin^2\theta$ can be written as $1 - $ theta.
An equation mixing sine and cosine is converted to one ratio using $\sin^2\theta = 1 - \cos^2\theta$, choosing the ratio that already appears unsquared. Then substitute and solve as a quadratic.
Pause, copy the conversion, the worked reduction to $2\cos^2\theta - 3\cos\theta + 1 = 0$, and the choose-the-unsquared-ratio rule, into your book.
Worked examples · 3 in a row, reveal as you go
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Work examples end to end
Follow the reasoning through complete worked solutions.
Run the quick drill and copy the summary into your book.
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Revisit the substitution you made
At the start you replaced $\sin\theta$ with $s$ in $2\sin^2\theta - \sin\theta - 1 = 0$. State the factorisation you found, and explain what still has to be done after solving for $s$ before the question is answered.
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Multiple choice
Answer the drill bank and rate your confidence.
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Multiple choice
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Short answer
Write full responses, then check them against the model answers.
Q2. Solve $2\sin^2\theta = 3\cos\theta$ for $0^\circ \leq \theta \leq 360^\circ$, showing how you reduce it to one ratio. (4 marks)
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UnderstandBand 42 marks
Q3. Explain why a root of $s = 1.4$ is rejected when $s = \sin\theta$, but a root of $t = 1.4$ is not rejected when $t = \tan\theta$. (2 marks)
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📖 Comprehensive answers (click to reveal)
Practice 1: $s^2 - s = 0$, so $s(s - 1) = 0$. Practice 2: cosine never exceeds $1$, so $\cos\theta = 3$ has no solutions. Practice 3: $\cos^2\theta = 1 - \sin^2\theta$. Practice 4: $\cos^2\theta = \dfrac{1}{2}$, so $\cos\theta = \pm\dfrac{\sqrt{2}}{2}$, giving $\theta = 45^\circ, 135^\circ$.
Q2 (4 marks): $\sin^2\theta = 1 - \cos^2\theta$, so $2 - 2\cos^2\theta = 3\cos\theta$ [1]. Rearranged, $2\cos^2\theta + 3\cos\theta - 2 = 0$, so $(2c - 1)(c + 2) = 0$ [1]. $c = -2$ is rejected, since cosine cannot be less than $-1$ [1]. $\cos\theta = \dfrac{1}{2}$ gives $\theta = 60^\circ, 300^\circ$ [1].
Q3 (2 marks): Sine is bounded: for any angle, $\sin\theta$ lies between $-1$ and $1$, so $\sin\theta = 1.4$ has no solutions and the root is rejected [1]. Tangent is unbounded and takes every real value, so $\tan\theta = 1.4$ does have solutions and the root is kept [1].
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Review and finish
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Boss battle · Quadratic Trig Trials
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Substitute, factorise, reject impossible roots, and solve every surviving equation on the domain. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.