Mathematics 9709 · AS & A Level · Trigonometry

Trigonometry — practice question

(i)[4]

Solve for $x$ in the equation $4\sin^2 x + 8\cos x - 7 = 0$ where $0^\circ \leq x \leq 360^\circ$.

(ii)[2]

Hence find the solution of the equation $4\sin^2\left(\frac{1}{2}\theta\right) + 8\cos\left(\frac{1}{2}\theta\right) - 7 = 0$ for $0^\circ \leq \theta \leq 360^\circ$.

Worked solution & mark scheme

This 6-mark question has a full step-by-step worked solution and mark scheme. One marking point: Form the equation $4(1-\cos^2 x)+8\cos x-7=0$.

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