Transformed argument solving

2 Find all the values of \(x\) in the interval \(0 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }\) which satisfy the equation \(\sin 3 x + 2 \cos 3 x = 0\).

1 Solve the equation \(3 \tan \left( 2 x + 15 ^ { \circ } \right) = 4\) for \(0 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }\).

4. Solve, for \(0 \leqslant x < 180 ^ { \circ }\), $$\cos \left( 3 x - 10 ^ { \circ } \right) = - 0.4$$ giving your answers to 1 decimal place. You should show each step in your working.

1. In this question you must show detailed reasoning.

Solutions relying entirely on calculator technology are not acceptable.

1. Show that the equation $$2 \tan \theta = 3 \cos \theta$$ can be written as $$3 \sin ^ { 2 } \theta + 2 \sin \theta - 3 = 0$$
2. Hence solve, for \(- \pi < x < \pi\), the equation $$2 \tan \left( 2 x + \frac { \pi } { 3 } \right) = 3 \cos \left( 2 x + \frac { \pi } { 3 } \right)$$ giving your answers to 3 significant figures.

5 Solve \(\sin \left( 2 \theta + 30 ^ { \circ } \right) = 0.1\) in the range \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).

3 Solve the equation \(\tan \left( \theta + 10 ^ { \circ } \right) = 0.1\) in the range \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

Find the exact solution of the equation $$\cos\frac{x}{6} + \tan 2x + \frac{\sqrt{3}}{2} = 0 \text{ for } -\frac{1}{4}\pi < x < \frac{1}{4}\pi.$$ [2]

Solve for \(0 \leq \theta \leq 180°\) $$\tan(\theta + 35°) = \cot(\theta - 53°)$$ [Total 4 marks]