Factorising Trigonometric Equations
Some trig equations refuse to yield to a single identity. The secret weapon? Treat them like algebra — take out a common factor, recognise a quadratic in disguise, or substitute $u = \sin x$. In this lesson you'll build the pattern-matching eye that lets you crack any factorisable trig equation in an HSC exam.
Look at $2\sin^2 x - \sin x - 1 = 0$. Without using any technique — what does this equation remind you of? Write your gut reaction below.
Every factorisable trig equation uses at least one of three moves. Recognise which applies, then reduce to simple equations.
Move 1 — Common factor: if every term shares a trig function, factor it out. E.g. $\sin x(\sin x - 1) = 0$.
Move 2 — Quadratic substitution: let $u = \sin x$ (or $\cos x$) to reveal a quadratic $au^2 + bu + c = 0$, factorise algebraically, then solve $u = k$.
Move 3 — Pythagorean identity: replace $\sin^2 x$ or $\cos^2 x$ using $\sin^2 x + \cos^2 x = 1$ to reduce to a single trig function before factorising.
Key facts
- Three factorisation strategies: common factor, quadratic substitution, Pythagorean identity
- $\sin^2 x + \cos^2 x = 1$ allows single-function reduction
- Zero-product property: $(A)(B) = 0 \Rightarrow A = 0$ or $B = 0$
Concepts
- Why factorising is preferable to dividing by a trig function
- How substitution transforms a trig equation into an algebraic one
- Why solutions outside $[-1, 1]$ must be rejected for $\sin$ and $\cos$
Skills
- Factor out a common trig factor and solve
- Use substitution to solve quadratic-type trig equations
- Apply a Pythagorean identity before factorising
When every term in a trig equation contains the same trig function, factor it out rather than dividing both sides. Division loses the solution where that function equals zero.
Strategy: Rearrange to get zero on one side, identify the common factor, apply the zero-product property.
Factor: $\cos x(\sin x - 1) = 0$
So $\cos x = 0$ or $\sin x = 1$.
$\cos x = 0$: $x = 90°, 270°$. $\sin x = 1$: $x = 90°$.
Combined (unique) solutions: $x = 90°, 270°$.
Notice: if we had instead divided both sides by $\cos x$, we would get $\sin x = 1$, losing $x = 270°$. Never divide by a trig expression that could be zero.
General form of common-factor equations:
- $\sin x \cdot f(\sin x, \cos x) = 0$
- $\cos x \cdot g(\sin x, \cos x) = 0$
- $\tan x \cdot h(x) = 0$
The common factor gives one set of solutions; the remaining factor gives the rest.
Move everything to one side: f(x) = 0, then factor; Never divide both sides by a trig function — factor instead
Pause — copy the factorisation rule into your book: always rearrange to $f(x) \cdot g(x) = 0$ then solve each factor separately; never divide by a trig function — that loses solutions where the function equals zero.
Quick check: Solving $2\sin x \cos x - \sin x = 0$ in $[0°, 360°)$, what is the correct first step?
We just saw that rearranging to $f(x) \cdot g(x) = 0$ and factoring prevents the loss of solutions that division causes. That raises a question: when the equation has both $\sin^2 x$ and $\sin x$ terms (a quadratic in disguise), what substitution converts it to a solvable polynomial? This card answers it → let $u = \sin x$ (or $u = \cos x$), solve the resulting quadratic, then back-substitute, rejecting any $|u| > 1$.
When the equation contains $\sin^2 x$ and $\sin x$ (or $\cos^2 x$ and $\cos x$), the equation is quadratic in disguise. The substitution $u = \sin x$ (or $u = \cos x$) reveals a standard algebraic quadratic.
Process:
- Rearrange so one side is zero.
- Let $u = \sin x$ (or $\cos x$); rewrite as $au^2 + bu + c = 0$.
- Factorise or use the quadratic formula to find $u$.
- Discard any $|u| > 1$ (out of range for $\sin$ or $\cos$).
- Solve $\sin x = u_1$ and $\sin x = u_2$ using CAST in the given domain.
Factorising: $(2u + 1)(u - 1) = 0$, so $u = -\tfrac{1}{2}$ or $u = 1$.
Both satisfy $|u| \leq 1$. Solve $\sin x = -\tfrac{1}{2}$ and $\sin x = 1$ in the given domain.
Let u = x or u = x when you see squared and linear terms of the same function; Always check -1 u 1 before solving each case
Pause — copy the quadratic-substitution rule into your book: let $u = \sin x$ (or $u = \cos x$) when the equation has both squared and linear trig terms; always reject solutions with $|u| > 1$ since they are outside the range.
Did you get this? True or false: when solving $2\cos^2 x + \cos x - 1 = 0$ using substitution $u = \cos x$, the factored form is $(2u - 1)(u + 1) = 0$.
Worked examples · 3 in a row, reveal as you go
Solve $\sin x \cos x = \sin x$ for $x \in [0°, 360°)$.
Solve $2\sin^2 x - \sin x - 1 = 0$ for $x \in [0°, 360°)$.
Solve $\cos^2 x - \sin x - 1 = 0$ for $x \in [0°, 360°)$.
Fill the gap: To solve $\cos^2 x + \cos x = 0$, factor as $\cos x(\cos x +$ $) = 0$ .
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: when solving $\cos^2 x - 3\cos x + 2 = 0$, the root $u = 2$ (i.e. $\cos x = 2$) should be rejected because it is outside the range of cosine.
Activities · practice with the ideas
Solve $\tan x \cdot \sin x = \tan x$ for $x \in [0°, 360°)$ by factoring out the common factor.
Solve $2\cos^2 x + \cos x - 1 = 0$ for $x \in [0°, 360°)$ using substitution.
Solve $\sin^2 x - \cos^2 x - \cos x = 0$ for $x \in [0°, 360°)$. (Hint: use a Pythagorean identity first.)
Solve $\cos^2 x - 3\cos x + 2 = 0$ for $x \in [0°, 360°)$. Identify any rejected roots clearly.
Show that the equation $\sin^2 x + 2\sin x + 2 = 0$ has no real solutions.
Odd one out: Three of these are valid first steps for solving $\sin^2 x - \sin x - 2 = 0$. Which step is WRONG?
Earlier you recorded your gut reaction to $2\sin^2 x - \sin x - 1 = 0$.
The solution is $x = 90°, 210°, 330°$ (for $[0°, 360°)$). Did you recognise the quadratic structure? The key insight is that $\sin^2 x$ and $\sin x$ behave just like $u^2$ and $u$ in algebra — once you spot this, every such equation becomes routine. The same moves you used in Year 10 algebra work here, applied to trig functions.
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 $\cos x(\cos x - 1) = 0$ for $x \in [0°, 360°)$. (2 marks)
Q2. Solve $2\sin^2 x + 3\sin x + 1 = 0$ for $x \in [0°, 360°)$. (3 marks)
Q3. Solve $\sin^2 x + \cos x + 1 = 0$ for $x \in [0°, 360°)$. Show all working. (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. $\tan x(\sin x - 1) = 0$: $\tan x = 0 \Rightarrow x = 0°, 180°$; $\sin x = 1 \Rightarrow x = 90°$. Solutions: $0°, 90°, 180°$.
2. Let $u = \cos x$: $(2u - 1)(u + 1) = 0 \Rightarrow u = \tfrac{1}{2}$ or $u = -1$. $\cos x = \tfrac{1}{2} \Rightarrow x = 60°, 300°$; $\cos x = -1 \Rightarrow x = 180°$. Solutions: $60°, 180°, 300°$.
3. $\sin^2 x = 1 - \cos^2 x$: $(1 - \cos^2 x) - \cos^2 x - \cos x = 0 \Rightarrow -2\cos^2 x - \cos x + 1 = 0 \Rightarrow 2\cos^2 x + \cos x - 1 = 0 \Rightarrow (2\cos x - 1)(\cos x + 1) = 0$. Solutions: $x = 60°, 180°, 300°$.
4. $(u-2)(u-1) = 0 \Rightarrow u = 2$ (reject, $|\cos x| \leq 1$) or $u = 1 \Rightarrow x = 0°$.
5. $u^2 + 2u + 2 = 0 \Rightarrow \Delta = 4 - 8 = -4 < 0$. No real roots $\Rightarrow$ no real solutions.
Q1 (2 marks): $\cos x = 0 \Rightarrow x = 90°, 270°$ [1]; $\cos x = 1 \Rightarrow x = 0°$ [1]. Solutions: $0°, 90°, 270°$.
Q2 (3 marks): $u = \sin x$: $(2u+1)(u+1)=0$ [1]; $u = -\tfrac{1}{2}$ or $u = -1$ — both valid [1]; $\sin x = -\tfrac{1}{2} \Rightarrow x = 210°, 330°$; $\sin x = -1 \Rightarrow x = 270°$ [1]. Solutions: $210°, 270°, 330°$.
Q3 (3 marks): $(1-\cos^2 x) + \cos x + 1 = 0 \Rightarrow -\cos^2 x + \cos x + 2 = 0 \Rightarrow \cos^2 x - \cos x - 2 = 0$ [1]; $(\cos x - 2)(\cos x + 1) = 0$ [1]; $\cos x = 2$ rejected; $\cos x = -1 \Rightarrow x = 180°$ [1].
Five timed questions on factorising 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 factorisation questions. Lighter alternative to the boss.
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