(i)[4]
First expand $\tan(2x + x)$ to show that the relation $\tan 3x = 3\cot x$ can be transformed into $\tan^4 x - 12\tan^2 x + 3 = 0$.
(ii)[3]
Hence solve $\tan 3x = 3\cot x$ for $0^{\circ} < x < 90^{\circ}$.
Mathematics 9709 · AS & A Level · Trigonometry
First expand $\tan(2x + x)$ to show that the relation $\tan 3x = 3\cot x$ can be transformed into $\tan^4 x - 12\tan^2 x + 3 = 0$.
Hence solve $\tan 3x = 3\cot x$ for $0^{\circ} < x < 90^{\circ}$.
This 7-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Apply the $\\tan(A+B)$ identity to write the LHS in terms of $\\tan 2x$ and $\\tan x$” …