Mathematics 9709 · AS & A Level · Trigonometry

Trigonometry — practice question

(a)[4]

Show that $\sqrt{5}\sec x + \tan x = 4$ may be written as $R\cos(x + \alpha) = \sqrt{5}$, with $R > 0$ and $0^\circ < \alpha < 90^\circ$. State the exact value of $R$ and give $\alpha$ correct to 2 decimal places.

(b)[4]

Hence determine the solutions of the equation $\sqrt{5}\sec 2x + \tan 2x = 4$, for $0^\circ < x < 180^\circ$.

Worked solution & mark scheme

This 8-mark question has a full step-by-step worked solution and mark scheme. One marking point: Manipulate the expression to reach $4\cos x-\sin x=\sqrt5$

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