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:
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 $R = \sqrt{a^2 + \square}$, where $\square =$ .
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)
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].
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