Product of trig functions

1 Solve the equation \(4 \sin \theta + \tan \theta = 0\) for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).

4 Find the values of \(\theta\) such that \(0 ^ { \circ } \leq \theta \leq 360 ^ { \circ }\) which satisfy the equation $$\cos \theta \tan \theta = \frac { \sqrt { 3 } } { 2 }$$

6 Solve the equation \(\tan \theta = 2 \sin \theta\) for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

7 Solve the equation \(\tan \theta = 2 \sin \theta\) for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

8 In this question you must show detailed reasoning.
Solve the equation \(3 \cos \theta + 8 \tan \theta = 0\) for \(0 ^ { \circ } < \theta < 360 ^ { \circ }\), giving your answers correct to the nearest degree.

1．Solve the equation \(\cos x + \sqrt { } \left( 1 - \frac { 1 } { 2 } \sin 2 x \right) = 0 , \quad\) in the interval \(0 ^ { \circ } \leq x < 360 ^ { \circ }\) ．

2．Solve，for \(0 \leqslant x \leqslant 360 ^ { \circ }\) $$\sin 47 ^ { \circ } \cos ^ { 3 } x + \cos 47 ^ { \circ } \sin x \cos ^ { 2 } x = \frac { 1 } { 2 } \cos ^ { 2 } x$$

7 Solve the equation $$\sin \theta \tan \theta + 2 \sin \theta = 3 \cos \theta \quad \text { where } \cos \theta \neq 0$$ Give all values of \(\theta\) to the nearest degree in the interval \(0 ^ { \circ } < \theta < 180 ^ { \circ }\) Fully justify your answer.
[0pt] [5 marks]