Trig equations, trig proofs and trig in context – A-level Maths exam questions with mark scheme

1CoreCalculator3 marks

Solve, for $0\leqslant x<360^\circ$, $$3\sin x=2\cos x,$$ giving your answers to 1 decimal place. [3]

for using $\tan x=\dfrac{\sin x}{\cos x}$ to obtain $\tan x=\tfrac23$M1
$33.7^\circ$ (awrt)A1
$213.7^\circ$ (awrt) – their $33.7^\circ+180^\circ$, and no extra solutions in rangeA1ft
Answer: $x=33.7^\circ,\ 213.7^\circ$
Examiner tip: Divide by $\cos x$, don't square both sides – squaring creates extra wrong solutions you then have to remove.

2StandardCalculator5 marks

Solve, for $0\leqslant\theta<360^\circ$, $$2\sin^{2}\theta+3\cos\theta=3.$$ [5]

for using $\sin^{2}\theta=1-\cos^{2}\theta$M1
for $2\cos^{2}\theta-3\cos\theta+1=0$ (or equivalent 3-term quadratic $=0$)A1
for solving their quadratic in $\cos\theta$: $(2\cos\theta-1)(\cos\theta-1)=0$M1
two of $0^\circ,\ 60^\circ,\ 300^\circ$A1
all three: $0^\circ,\ 60^\circ,\ 300^\circ$ and no others in rangeA1
Answer: $\theta=0^\circ,\ 60^\circ,\ 300^\circ$
Examiner tip: $\cos\theta=1$ gives $\theta=0^\circ$ – it's in the interval because of '$\leqslant$', and it's the solution most often missed.

3HardCalculator4 marks

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

Solve, for $0\leqslant x<180^\circ$, $$\tan(2x-30^\circ)=-\sqrt{3}.$$ [4]

for $2x-30^\circ=120^\circ$ (or $-60^\circ$), i.e. a correct order of operations with the exact valueM1
$x=75^\circ$A1
for adding $180^\circ$ to their $2x-30^\circ$ value to find a second value in the transformed range $-30^\circ\leqslant 2x-30^\circ<330^\circ$M1
$x=165^\circ$ and no others in rangeA1
Answer: $x=75^\circ,\ 165^\circ$
Examiner tip: Change the interval first: $-30^\circ\leqslant 2x-30^\circ<330^\circ$; find all angles there, then add 30 and halve.

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