Solve equation with double angle

2 Solve the equation $$\sin 2 x + 3 \cos 2 x = 0$$ for \(0 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }\).

2 Solve the equation \(3 \sin 2 \theta \tan \theta = 2\) for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).

3 Solve the equation \(\tan 2 x = 5 \cot x\), for \(0 ^ { \circ } < x < 180 ^ { \circ }\).

3 Showing all necessary working, solve the equation \(\cot 2 \theta = 2 \tan \theta\) for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).

2 Solve the equation $$\tan x \tan 2 x = 1 ,$$ giving all solutions in the interval \(0 ^ { \circ } < x < 180 ^ { \circ }\).

2 Solve the equation \(5 \cos \theta ( 1 + \cos 2 \theta ) = 4\) for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

2 Solve the equation \(3 \cos 2 \theta = 3 \cos \theta + 2\), for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

5 Solve the equation \(\sin \theta = 3 \cos 2 \theta + 2\), for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

1. Solve, for \(0 \leqslant x < 360 ^ { \circ }\), the equation

$$2 \cos 2 x = 7 \cos x$$ giving your solutions to one decimal place.
(Solutions based entirely on graphical or numerical methods are not acceptable.)

6. (i) Without using a calculator, find the exact value of $$\left( \sin 22.5 ^ { \circ } + \cos 22.5 ^ { \circ } \right) ^ { 2 }$$ You must show each stage of your working.
(ii) (a) Show that \(\cos 2 \theta + \sin \theta = 1\) may be written in the form $$k \sin ^ { 2 } \theta - \sin \theta = 0 , \text { stating the value of } k$$ (b) Hence solve, for \(0 \leqslant \theta < 360 ^ { \circ }\), the equation $$\cos 2 \theta + \sin \theta = 1$$

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 \(0 \leqslant \theta < \pi\), solve $$2 \sin 2 \theta = 3 ( \cos 2 \theta + \sin \theta - 1 )$$ giving your answers in radians to 3 significant figures, where appropriate.

8. (a) Write down \(\sin 2 x\) in terms of \(\sin x\) and \(\cos x\).
(b) Find, for \(0 < x < \pi\), all the solutions of the equation $$\operatorname { cosec } x - 8 \cos x = 0$$ giving your answers to 2 decimal places.

1. (i) Find the exact value of \(x\) such that

$$3 \tan ^ { - 1 } ( x - 2 ) + \pi = 0$$ (ii) Solve, for \(- \pi < \theta < \pi\), the equation $$\cos 2 \theta - \sin \theta - 1 = 0$$ giving your answers in terms of \(\pi\).

4 Solve the equation \(2 \sin 2 \theta = 1 + \cos 2 \theta\) for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).

1 Solve the equation \(2 \sin 2 \theta = \cos \theta\) for \(0 ^ { \circ } \leq \theta \leq 360 ^ { \circ }\).

4 Solve the equation \(\cos 2 \theta = \sin \theta\) for \(0 \leqslant \theta \leqslant 2 \pi\), giving your answers in terms of \(\pi\).

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

6 Express \(6 \cos 2 \theta + \sin \theta\) in terms of \(\sin \theta\).
Hence solve the equation \(6 \cos 2 \theta + \sin \theta = 0\), for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

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

3．Solve for values of \(\theta\) ，in degrees，in the range \(0 \leq \theta \leq 360\) ， $$\sqrt { } 2 \times ( \sin 2 \theta + \cos \theta ) + \cos 3 \theta = \sin 2 \theta + \cos \theta$$

1．Solve the following equation，for \(0 \leq x \leq \pi\) ，giving your answers in terms of \(\pi\) ． $$\sin 5 x - \cos 5 x = \cos x - \sin x$$

2．Solve，for \(0 < \theta < 2 \pi\) ， $$\sin 2 \theta + \cos 2 \theta + 1 = \sqrt { 6 } \cos \theta$$ giving your answers in terms of \(\pi\) ．

2．Given that \(( \sin \theta + \cos \theta ) \neq 0\) ，find all the solutions of $$\frac { 2 \cos 2 \theta ( \sin 2 \theta - \sqrt { } 3 \cos 2 \theta ) } { \sin \theta + \cos \theta } = \sqrt { } 6 ( \sin 2 \theta - \sqrt { } 3 \cos 2 \theta )$$ for \(0 \leq \theta < 360 ^ { \circ }\) ．

3．（a）Solve，for \(0 \leq x < 2 \pi\) ， $$\cos x + \cos 2 x = 0$$ （b）Find the exact value of \(x , x \geq 0\) ，for which $$\arccos x + \arccos 2 x = \frac { \pi } { 2 }$$ ［ \(\arccos x\) is an alternative notation for \(\cos ^ { - 1 } x\) ．］

2 Express \(6 \cos 2 \theta + \sin \theta\) in terms of \(\sin \theta\).
Hence solve the equation \(6 \cos 2 \theta + \sin \theta = 0\), for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).