Mixed Trigonometric Problem Solving
HSC questions rarely label themselves "auxiliary angle" or "t-formula" — you have to read the equation, choose the right tool, and execute. This lesson is your master class in technique selection: factorising, substitution, auxiliary angle, t-formula, and general solutions all in one session. Break the problem down, pick the sharpest tool, and verify.
$3\sin x - \cos x = 1$. Before doing any working — which technique would you reach for first? Why that one and not another? Write your intuition below.
Read the equation structure, then pick the sharpest matching tool:
- $a\sin x + b\cos x = c$ — try auxiliary angle or t-formula
- product of trig functions = 0 — try factorisation, then solve each factor
- quadratic in $\sin x$ or $\cos x$ — try substitution $u=\sin x$ (or $\cos x$), then factor/quadratic formula
- double/multiple angle present — use double-angle identities to reduce to single angle
- need all solutions on $\mathbb{R}$ or in wide interval — write the general solution
Key facts
- Which equation types call for auxiliary angle, t-formula, factorisation, substitution, or general solution
- The double-angle identities: $\sin 2x = 2\sin x\cos x$; $\cos 2x = 1-2\sin^2 x = 2\cos^2 x - 1$
- The t-formula substitutions: $\sin x = \tfrac{2t}{1+t^2}$, $\cos x = \tfrac{1-t^2}{1+t^2}$, $t=\tan\tfrac{x}{2}$
Concepts
- Why a linear combination $a\sin x + b\cos x$ has amplitude $\sqrt{a^2+b^2}$
- Why the t-formula can miss solutions at $x = \pi + 2k\pi$ (where $\cos x = -1$)
- How factorisation decomposes a product equation into simpler cases
Skills
- Classify any trig equation and select the best technique
- Solve multi-step problems combining two or more techniques
- Verify solutions and identify/remove extraneous ones
When the equation can be rearranged to a product equal to zero, set each factor to zero and solve independently.
Example: Solve $\sin x \cos x = \tfrac{1}{2}\sin x$ for $0\le x\le 2\pi$.
Factor 1: $\sin x = 0 \Rightarrow x = 0, \pi, 2\pi$. Factor 2: $\cos x = \tfrac{1}{2} \Rightarrow x = \tfrac{\pi}{3}, \tfrac{5\pi}{3}$.
Solutions: $x = 0, \tfrac{\pi}{3}, \pi, \tfrac{5\pi}{3}, 2\pi$.
Rearrange to f(x) g(x) = 0, then solve f(x)=0 and g(x)=0 separately; Never divide by a trig expression — factor it out instead
Pause — copy the factorisation technique into your book: rearrange to $f(x)\cdot g(x) = 0$, solve each factor separately; never cancel a trig factor from both sides — always factor it out to avoid losing solutions.
Quick check: Which first step is best for $2\sin^2 x + \sin x - 1 = 0$?
We just saw that multi-technique questions require identifying the equation type first — factorisation, quadratic substitution, or auxiliary angle — then executing the matching procedure. That raises a question: when an equation needs both the auxiliary angle conversion and factorisation, in which order do you apply them? This card answers it → convert to $R\sin(x+\phi)$ first to collect the $\sin$ and $\cos$ terms, then rearrange and factorise the result.
For $a\sin x + b\cos x = c$, convert to $R\sin(x + \varphi) = c$ where $R = \sqrt{a^2+b^2}$ and $\tan\varphi = b/a$ (with $\varphi$ in the correct quadrant).
Example: Solve $3\sin x - \cos x = 1$ for $0\le x\le 2\pi$.
$R = \sqrt{9+1} = \sqrt{10}$. Write $3\sin x - \cos x = \sqrt{10}\sin(x-\varphi)$ where $\tan\varphi = \tfrac{1}{3}$, $\varphi = \arctan\tfrac{1}{3}$.
Let $\beta = \sin^{-1}\!\tfrac{1}{\sqrt{10}}$. Solutions for $x-\varphi$: $\beta$ and $\pi-\beta$ (in $[0,2\pi]$). Then $x = \varphi+\beta$ and $x = \varphi+\pi-\beta$ (adjust for interval).
a x + b x = R(x+): R=a^2+b^2, = b/a; Check: R = a and R = b
Pause — copy the auxiliary angle formula into your book: $a\sin x + b\cos x = R\sin(x+\varphi)$ where $R = \sqrt{a^2+b^2}$; verify with $R\cos\varphi = a$ and $R\sin\varphi = b$.
Did you get this? True or false: the equation $2\sin x + 2\cos x = 5$ has no real solutions.
Worked examples · 3 in a row, reveal as you go
Solve $\cos 2x + \cos x = 0$ for $0 \le x \le 2\pi$.
Solve $3\cos x + \sin x = 2$ for $0 \le x \le 2\pi$ using the t-formula.
Solve $2\sin^2 x + \sin x - 1 = 0$ for $0 \le x \le 2\pi$.
Fill the gap: For $a\sin x + b\cos x = R\sin(x+\varphi)$, the amplitude is $R = \sqrt{a^2 +$ $}$.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: when solving $\sin x\tan x = \sin x$, dividing both sides by $\sin x$ gives all solutions.
Activities · practice with the ideas
Solve $\sin x\cos x = \cos x$ for $0 \le x \le 2\pi$ by factorisation.
Write $\sqrt{3}\sin x + \cos x$ in the form $R\sin(x+\varphi)$ and hence solve $\sqrt{3}\sin x + \cos x = 1$ for $0\le x\le 2\pi$.
Solve $\cos 2x - 3\cos x + 2 = 0$ for $0\le x\le 2\pi$. (Hint: choose the right double-angle form.)
Solve $2\cos^2 x + 3\sin x - 3 = 0$ for $0\le x\le 2\pi$. (Hint: use $\cos^2 x = 1 - \sin^2 x$.)
Use the t-formula to solve $\sin x - \cos x = 1$ for $0\le x\le 2\pi$. Check for solutions at $x=\pi$.
Odd one out: Three of these are valid first moves for $\sin 2x - \sin x = 0$. Which one is NOT appropriate as a first step?
Earlier you predicted which technique to use for $3\sin x - \cos x = 1$.
Both auxiliary angle and t-formula work here. Auxiliary angle converts to $\sqrt{10}\sin(x-\varphi)=1$ (elegant and exact). The t-formula gives a quadratic in $t$ that needs a calculator for exact values. The key lesson: recognise the "$a\sin x + b\cos x$" pattern and reach for auxiliary angle first — it's cleaner.
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 $\sin x(\sin x - 1) = 0$ for $0 \le x \le 2\pi$. (2 marks)
Q2. Solve $\cos 2x = \cos x$ for $0 \le x \le 2\pi$. Show all working. (3 marks)
Q3. Express $\sin x + \sqrt{3}\cos x$ in the form $R\sin(x+\varphi)$, and hence find all solutions of $\sin x + \sqrt{3}\cos x = \sqrt{3}$ for $0 \le x \le 2\pi$. (4 marks)
Comprehensive answers (click to reveal)
Activity answers: 1. $\sin x(\cos x-1)=0$: $\sin x=0\Rightarrow x=0,\pi,2\pi$; $\cos x=1\Rightarrow x=0,2\pi$; combined: $x=0,\pi,2\pi$ · 2. $2\sin(x+\tfrac{\pi}{6})=1\Rightarrow\sin(x+\tfrac{\pi}{6})=\tfrac{1}{2}$; $x+\tfrac{\pi}{6}=\tfrac{\pi}{6},\tfrac{5\pi}{6}$; $x=0,\tfrac{2\pi}{3}$ · 3. $\cos 2x=2\cos^2 x-1$; $2\cos^2 x-3\cos x+1=0$; $(2\cos x-1)(\cos x-1)=0$; $x=\tfrac{\pi}{3},\tfrac{5\pi}{3},0,2\pi$ · 4. $\cos^2 x=1-\sin^2 x$; $-2\sin^2 x+3\sin x-1=0$; $2\sin^2 x-3\sin x+1=0$; $(2\sin x-1)(\sin x-1)=0$; $x=\tfrac{\pi}{6},\tfrac{5\pi}{6},\tfrac{\pi}{2}$ · 5. t-formula: $t^2-t=0\Rightarrow t(t-1)=0$; $t=0\Rightarrow x=0,2\pi$; $t=1\Rightarrow x=\tfrac{\pi}{2}$; check $x=\pi$: $\sin\pi-\cos\pi=0-(-1)=1$ ✓ — include $x=\pi$; solutions: $x=0,\tfrac{\pi}{2},\pi,2\pi$
Q1 (2 marks): $\sin x=0\Rightarrow x=0,\pi,2\pi$ [1]; $\sin x=1\Rightarrow x=\tfrac{\pi}{2}$ [1]. Solutions: $x=0,\tfrac{\pi}{2},\pi,2\pi$.
Q2 (3 marks): $\cos 2x = 2\cos^2 x-1$; equation: $2\cos^2 x - \cos x - 1 = 0$ [1]. $(2\cos x+1)(\cos x-1)=0$: $\cos x=-\tfrac{1}{2}$ or $\cos x=1$ [1]. $\cos x=-\tfrac{1}{2}\Rightarrow x=\tfrac{2\pi}{3},\tfrac{4\pi}{3}$; $\cos x=1\Rightarrow x=0,2\pi$ [1].
Q3 (4 marks): $R=\sqrt{1+3}=2$ [1]; $\tan\varphi=\sqrt{3}\Rightarrow\varphi=\tfrac{\pi}{3}$ (Q1, both positive) [1]. $2\sin(x+\tfrac{\pi}{3})=\sqrt{3}\Rightarrow\sin(x+\tfrac{\pi}{3})=\tfrac{\sqrt{3}}{2}$ [1]. $x+\tfrac{\pi}{3}=\tfrac{\pi}{3},\tfrac{2\pi}{3}$; $x=0,\tfrac{\pi}{3}$ — check full range: $x+\tfrac{\pi}{3}\in[\tfrac{\pi}{3},\tfrac{7\pi}{3}]$; also $x+\tfrac{\pi}{3}=\pi+\tfrac{\pi}{3}=\tfrac{4\pi}{3}$ (sin=−, invalid); so solutions $x=0,\tfrac{\pi}{3}$ [1].
Five timed questions combining all Module 7 techniques. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering mixed-technique trig questions. Lighter alternative to the boss.
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