Solve equation with double angle

5 Solve the equation \(2 \cos 2 x = 1 + \cos x\), for \(0 ^ { \circ } \leqslant x < 360 ^ { \circ }\).

6. Solve the equation $$\cos 2 \theta - \cos 4 \theta = \sin 3 \theta \quad \text { for } \quad 0 \leqslant \theta \leqslant \pi$$

12. Find the general solution of the equation $$\cos 4 \theta + \cos 2 \theta = \cos \theta$$

9. Find the general solution of the equation $$\sin 6 \theta + 2 \cos ^ { 2 } \theta = 3 \cos 2 \theta - \sin 2 \theta + 1 .$$

5. (a) Using the identity \(\cos ( A + B ) \equiv \cos A \cos B - \sin A \sin B\), prove that $$\cos 2 A \equiv 1 - 2 \sin ^ { 2 } A$$ (b) Show that $$2 \sin 2 \theta - 3 \cos 2 \theta - 3 \sin \theta + 3 \equiv \sin \theta ( 4 \cos \theta + 6 \sin \theta - 3 )$$ (c) Express \(4 \cos \theta + 6 \sin \theta\) in the form \(R \sin ( \theta + \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { 1 } { 2 } \pi\).
(d) Hence, for \(0 \leq \theta < \pi\), solve $$2 \sin 2 \theta = 3 ( \cos 2 \theta + \sin \theta - 1 )$$ giving your answers in radians to 3 significant figures, where appropriate.
Hence, for \(0 \leq \theta < \pi\), solve
\includegraphics[max width=\textwidth, alt={}]{933ec0b9-3496-455e-9c2c-2612e84f63ff-02_20_26_1509_239} giving your answers in radians to 3 significant figures, where appropriate.

3

1. Express \(\cos 2 x\) in terms of \(\sin x\).
2. 1. Hence show that \(3 \sin x - \cos 2 x = 2 \sin ^ { 2 } x + 3 \sin x - 1\) for all values of \(x\).
   2. Solve the equation \(3 \sin x - \cos 2 x = 1\) for \(0 ^ { \circ } < x < 360 ^ { \circ }\).
3. Use your answer from part (a) to find \(\int \sin ^ { 2 } x \mathrm {~d} x\).

5

1. Express \(\sin 2 x\) in terms of \(\sin x\) and \(\cos x\).
2. Solve the equation $$5 \sin 2 x + 3 \cos x = 0$$ giving all solutions in the interval \(0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }\) to the nearest \(0.1 ^ { \circ }\), where appropriate.
3. Given that \(\sin 2 x + \cos 2 x = 1 + \sin x\) and \(\sin x \neq 0\), show that \(2 ( \cos x - \sin x ) = 1\).