Show that 2−2cos2x2+2cos2x≡tan2x\displaystyle \frac{2 - 2\cos 2x}{2 + 2\cos 2x} \equiv \tan^2 x2+2cos2x2−2cos2x≡tan2x
Hence solve, for −π≤x≤π-\pi \leq x \leq \pi−π≤x≤π,
2−2cos2x2+2cos2x=13 \frac{2 - 2\cos 2x}{2 + 2\cos 2x} = \frac{1}{3} 2+2cos2x2−2cos2x=31348 exam-style questions on Edexcel A Level Maths Trigonometry and Modelling, covering 7.1 Addition Formulae, 7.2 Double Angle Formulae, 7.3 Solving Trigonometric Equations, 7.4 Simplifying a cos x +- b sin x, 7.5 Proving Trigonometric Identities, and 7.6 Modelling with Trigonometric Functions. Each one has a worked solution and a mark scheme showing where the marks go.