Combining Techniques
The hardest HSC trig equations don't announce which method to use. Some need the auxiliary angle form first, then factorisation. Others need a Pythagorean identity, then a t-formula, then a general-solution argument. In this lesson you'll develop the strategic thinking to read an equation and build a route to the answer.
Consider $\sqrt{3}\sin x - \cos x = 1$ for $x \in [0°, 360°)$. Before reading on — which of auxiliary angle, factorisation, t-formula, or identity would you try first, and why?
When you face a trig equation you haven't seen before, run through this four-question checklist in order:
Q1 — Is it of the form $a\sin x + b\cos x = c$? → Auxiliary angle (or t-formula if allowed)
Q2 — Does a common trig factor divide every term? → Factor it out (never cancel)
Q3 — Are $\sin^2 x$ and $\sin x$ (or $\cos^2 x$ and $\cos x$) both present? → Quadratic substitution
Q4 — Are two different trig functions mixed (e.g. $\sin^2 x$ and $\cos x$)? → Apply a Pythagorean identity first, then revisit Q1–Q3
Key facts
- A four-question diagnostic checklist for choosing a technique
- The auxiliary-angle form $R\sin(x + \alpha)$ for $a\sin x + b\cos x$
- How factorisation and identity application can follow an auxiliary conversion
Concepts
- Why the order of technique application matters for multi-step equations
- How different techniques each simplify the equation structure one step at a time
- Why verifying in the original equation catches errors from combined methods
Skills
- Diagnose an unfamiliar trig equation and choose the right first step
- Chain two or more techniques (e.g. auxiliary → factorise, identity → substitute → CAST)
- Present a coherent multi-step solution with full working
To solve $a\sin x + b\cos x = c$, convert the left side to a single sinusoid:
The equation becomes $R\sin(x + \alpha) = c$, which reduces to $\sin\theta = \dfrac{c}{R}$ where $\theta = x + \alpha$.
When combining with factorisation: Sometimes after reaching $R\sin(x+\alpha) = c$, the value $\dfrac{c}{R}$ yields a "nice" angle. Alternatively, if $c = 0$, then $\sin(x+\alpha) = 0$ and you solve by factorisation or directly. Complexity arises when the post-auxiliary equation still has structure — e.g., is a quadratic in $\sin(x+\alpha)$.
Here $a = \sqrt{3}$, $b = -1$. $R = \sqrt{3+1} = 2$. $\tan\alpha = \dfrac{-1}{\sqrt{3}} \Rightarrow \alpha = -30°$ (or equivalently, write as $R\sin(x - 30°)$).
Equation: $2\sin(x - 30°) = 1 \Rightarrow \sin(x-30°) = \tfrac{1}{2}$.
Reference angle: $30°$. $\sin$ positive in Q1, Q2: $x - 30° = 30°$ or $x - 30° = 150°$.
Solutions: $x = 60°$ or $x = 180°$.
a x + b x = R(x+) where R = a^2+b^2, = b/a; After converting: solve (x+) = c/R (note domain adjustment for x+)
Pause — copy the auxiliary angle recap into your book: $a\sin x + b\cos x = R\sin(x+\phi)$ where $R = \sqrt{a^2+b^2}$; after converting, the domain for $(x+\phi)$ shifts by $\phi$ from the original domain of $x$.
Quick check: For $\sqrt{3}\sin x + \cos x$, what is the value of $R$?
We just saw the auxiliary angle method applied to equations — convert to $R\sin(x+\phi)$, then solve by dividing and shifting the domain. That raises a question: what if the equation contains a mix of $\sin$ and $\cos$ terms that can't be grouped into a single $a\sin x + b\cos x$ — should you apply an identity first instead? This card answers it → yes: apply a Pythagorean or double-angle identity to reduce to one trig function, then factorise before applying ASTC.
Many multi-technique problems follow this two-step chain:
- Apply a Pythagorean identity to write the equation entirely in one trig function.
- Factorise (common factor or quadratic substitution) and solve each case.
Example: Solve $2\cos^2 x - \sin x = 1$ for $x \in [0°, 360°)$.
Step 1 — Replace $\cos^2 x$: $2(1-\sin^2 x) - \sin x = 1 \Rightarrow 2 - 2\sin^2 x - \sin x = 1$
Step 2 — Rearrange: $2\sin^2 x + \sin x - 1 = 0$
Step 3 — Quadratic substitution ($u = \sin x$): $(2u - 1)(u + 1) = 0 \Rightarrow u = \tfrac{1}{2}$ or $u = -1$
Step 4 — CAST: $\sin x = \tfrac{1}{2} \Rightarrow x = 30°, 150°$; $\sin x = -1 \Rightarrow x = 270°$
Solutions: $x = 30°, 150°, 270°$.
Chain: identity → single function → factorise → CAST; Always rearrange to zero before factorising
Pause — copy the identity-then-factorisation chain into your book: apply Pythagorean or double-angle identity → reduce to one trig function → rearrange to zero → factorise → apply ASTC.
Did you get this? True or false: when solving $2\cos^2 x - \sin x = 1$, the correct identity to apply first is $\cos^2 x = 1 - \sin^2 x$.
Worked examples · 3 in a row, reveal as you go
Solve $\sin x + \cos x = 1$ for $x \in [0°, 360°)$ using the auxiliary angle method.
Solve $\sin^2 x - \cos x - 1 = 0$ for $x \in [0°, 360°)$.
Solve $2\sin x\cos x + \cos x - 2\sin x - 1 = 0$ for $x \in [0°, 360°)$.
Fill the gap: For $\sin x + \cos x$, the auxiliary angle form is $\sqrt{2}\sin(x +$ $°)$ .
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: when solving $\sin x + \cos x = 1$ by auxiliary angle, the domain for $\theta = x + 45°$ becomes $[45°, 405°)$, not $[0°, 360°)$.
Activities · practice with the ideas
Solve $\sin x - \cos x = 1$ for $x \in [0°, 360°)$ using auxiliary angle. Remember to adjust the domain for $\theta = x - 45°$.
Solve $2\cos^2 x - 3\cos x + 1 = 0$ for $x \in [0°, 360°)$ using quadratic substitution. Identify any rejected roots.
Solve $3\sin^2 x - 2\cos x - 2 = 0$ for $x \in [0°, 360°)$. (Hint: apply a Pythagorean identity first.)
Solve $\sin 2x + \sin x = 0$ for $x \in [0°, 360°)$. (Hint: replace $\sin 2x$ with the double-angle identity, then factor.)
Solve $\cos 2x + \cos x = 0$ for $x \in [0°, 360°)$ using the double-angle identity $\cos 2x = 2\cos^2 x - 1$.
Odd one out: Three of the following are valid first steps for solving $3\sin x - \sqrt{3}\cos x = \sqrt{3}$. Which one is WRONG?
At the start you recorded your instinct for solving $\sqrt{3}\sin x - \cos x = 1$.
The answer is $x = 60°$ or $x = 180°$, found via auxiliary angle: $R = 2$, $\alpha = -30°$, giving $\sin(x-30°) = \tfrac{1}{2}$. The key insight is that recognising the $a\sin x + b\cos x$ pattern instantly tells you which method to use — and the domain shift for $\theta = x - 30°$ (from $[-30°, 330°)$) avoids missing the second solution.
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 $\sin x + \cos x = 0$ for $x \in [0°, 360°)$. (2 marks)
Q2. Solve $\cos 2x + \cos x = 0$ for $x \in [0°, 360°)$, using $\cos 2x = 2\cos^2 x - 1$. (3 marks)
Q3. Solve $\sin x + \cos x = \sqrt{2}\sin x \cos x$ for $x \in [0°, 360°)$. Show full working and state any rejected cases. (4 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. $\sqrt{2}\sin(x-45°) = 1$; $\sin(x-45°) = \tfrac{1}{\sqrt{2}}$; domain $[-45°, 315°)$; $x-45° = 45°$ or $x-45° = 135°$. Solutions: $x = 90°, 180°$.
2. $(2u-1)(u-1)=0$; $u=\tfrac{1}{2}$ or $u=1$; both valid. $\cos x = \tfrac{1}{2} \Rightarrow x=60°, 300°$; $\cos x = 1 \Rightarrow x=0°$.
3. $3(1-\cos^2 x) - 2\cos x - 2 = 0 \Rightarrow -3\cos^2 x - 2\cos x + 1 = 0 \Rightarrow 3\cos^2 x + 2\cos x - 1 = 0 \Rightarrow (3\cos x - 1)(\cos x + 1) = 0$. $\cos x = \tfrac{1}{3} \Rightarrow x \approx 70.5°, 289.5°$; $\cos x = -1 \Rightarrow x = 180°$.
4. $2\sin x\cos x + \sin x = 0 \Rightarrow \sin x(2\cos x + 1) = 0$. $\sin x = 0 \Rightarrow x = 0°, 180°$; $\cos x = -\tfrac{1}{2} \Rightarrow x = 120°, 240°$.
5. $2\cos^2 x - 1 + \cos x = 0 \Rightarrow (2\cos x - 1)(\cos x + 1) = 0$. $\cos x = \tfrac{1}{2} \Rightarrow x = 60°, 300°$; $\cos x = -1 \Rightarrow x = 180°$.
Q1 (2 marks): $\sin x = -\cos x \Rightarrow \tan x = -1$ [1]. $x = 135°, 315°$ [1].
Q2 (3 marks): $(2\cos^2 x - 1) + \cos x = 0 \Rightarrow 2\cos^2 x + \cos x - 1 = 0$ [1]; $(2\cos x - 1)(\cos x + 1) = 0$ [1]; $x = 60°, 180°, 300°$ [1].
Q3 (4 marks): Note $\sin x\cos x = \tfrac{1}{2}\sin 2x$. Use auxiliary: LHS $= \sqrt{2}\sin(x+45°)$. RHS $= \tfrac{\sqrt{2}}{2}\sin 2x = \tfrac{\sqrt{2}}{2}(2\sin x\cos x)$. Alternatively: $\sin x + \cos x = \sqrt{2}\sin x\cos x$; square both sides: $(1 + 2\sin x\cos x) = 2\sin^2 x\cos^2 x$; let $s = \sin x\cos x = \tfrac{1}{2}\sin 2x$: $1 + 2s = 2s^2 \Rightarrow 2s^2 - 2s - 1 = 0 \Rightarrow s = \tfrac{2 \pm \sqrt{4+8}}{4} = \tfrac{1 \pm \sqrt{3}}{2}$. [1 each step, verify by substitution for final mark.] This is a Band 5–6 extension — any clear, systematic approach with valid working earns method marks.
Five timed questions combining all techniques from Module 7. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering combined-technique questions. Lighter alternative to the boss.
Mark lesson as complete
Tick when you've finished the practice and review.