###### 10.1

Given the equation $$7\cos x + 5 = 2\sin^2 x$$.

Solve it for $$0^\circ \leqslant x \leqslant 180^\circ$$. Give your answer in degrees.

Since $$\sin^2 x + \cos^2 x = 1$$, we obtain

$$7\cos x + 5 = 2\sin^2x\\ 7\cos x + 5 = 2(1-\cos^2x)\\ 2\cos^2x+7\cos x+3=0$$

It's a quadratic equation for a variable $$t=\cos x$$, so

$$D=b^2-4ac=7^2-4\times2\times3 = 49 - 24 = 25$$

$$\cos x = {-b \pm \sqrt{D} \over 2a} = \frac{-7\pm5}{4}$$

Therefore, $$\cos x = -\frac12$$ or $$\cos x =-3$$. Since $$|\cos x|\leqslant 1$$, the second value is not suitable. The only solution to the equation $$\cos x = -\frac12$$ for $$0^\circ \leqslant x \leqslant 180^\circ$$ is $$x=120^\circ$$.

###### 10.2

Given the equation $$4\sin^2x=3$$.

Solve it for $$-90^\circ \leqslant x \leqslant 0^\circ$$. Give your answer in degrees.

$$4\sin^2x=3\\ \sin^2x=\frac34\\ \sin x = \pm\sqrt{\frac34}\\ \sin x = \pm \frac{\sqrt3}{2}$$

The only root of the equation, satisfying $$-90^\circ \leqslant x \leqslant 0^\circ$$, is $$x=-60^\circ$$.

###### 10.3

Given the equation $$\sqrt3\sin2x+\cos2x=0$$.

Solve it for $$0^\circ \leqslant x \leqslant 90^\circ$$. Give your answer in degrees.

$$\sqrt3\sin2x+\cos2x = 2(\frac{\sqrt3}2\sin2x+\frac12\cos2x) = 2(\cos30^\circ\sin2x+\sin30^\circ\cos2x) = 2\sin(2x+30^\circ)$$

So, the equation $$\sqrt3\sin2x+\cos2x$$ is equivalent to

$$2\sin(2x+30^\circ) = 0\\ \sin(2x+30^\circ) = 0\\ 2x+30^\circ=180^\circ k,\ k\in\mathbb Z\\ 2x = 180^\circ k - 30^\circ\\ x = 90^\circ k - 15^\circ$$

The solution satisfies $$0^\circ \leqslant x \leqslant 90^\circ$$ when $$k=1$$, so $$x = 90^\circ - 15^\circ = 75^\circ$$.

###### 10.4

Given the equation $$2\cos^2\theta=3\sin\theta$$.

Solve it for $$-90^\circ \leqslant \theta \leqslant 90^\circ$$. Give your answer in degrees.

Since $$\sin^2 \theta + \cos^2 \theta = 1$$, we obtain

$$2\cos^2\theta=3\sin\theta\\ 2(1-\sin^2\theta) = 3\sin \theta\\ 2\sin^2\theta+3\sin\theta-2=0$$

It's a quadratic equation for a variable $$t=\sin \theta$$, so

$$D=b^2-4ac=3^2-4\times2\times(-2) = 9 + 16 = 25$$

$$\sin\theta = {-b \pm \sqrt{D} \over 2a} = \frac{-3\pm5}{4}$$

Therefore, $$\sin \theta = \frac12$$ or $$\sin \theta = -2$$. Since $$|\sin\theta|\leqslant 1$$, the second value is not suitable. The only solution to the equation $$\sin \theta = \frac12$$ for $$-90^\circ \leqslant \theta \leqslant 90^\circ$$ is $$\theta=30^\circ$$.

###### 10.5

Given that $$\tan (\alpha + \beta) = 3$$ and $$\tan \beta = -1$$.

Find $$\tan \alpha$$.

It's known that $$\tan (\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1-\tan\alpha\tan\beta}$$, so

$$3 = \frac{\tan\alpha - 1}{1+\tan\alpha}\\ 3(1+\tan\alpha) = \tan\alpha -1\\ 2\tan\alpha = -4\\ \tan\alpha = -2$$

###### 10.6

Given the equation $$2\cos2\alpha = 4\cos\alpha-3$$.

Solve it for $$180^\circ \leqslant \alpha \leqslant 360^\circ$$. Give your answer in degrees.

Since $$\cos 2\alpha = 2 \cos^2\alpha-1$$, we obtain

$$2\cos2\alpha = 4\cos\alpha-3\\ 2(2 \cos^2\alpha-1) = 4 \cos\alpha -3\\ 4\cos^2\alpha-4\cos\alpha+1=0\\ (2\cos\alpha-1)^2=0\\ 2\cos\alpha-1 = 0\\ \cos\alpha = \frac12$$

The only solution to that equation for $$180^\circ \leqslant \alpha \leqslant 360^\circ$$ is $$\alpha = 300^\circ$$.

###### 10.7

Given the equation $$\sqrt3\sin^2x - \sin 2x = \sqrt3\cos^2x$$.

Solve it for $$0^\circ \leqslant x \leqslant 180^\circ$$. Give your answer in degrees.

Since $$\cos^2 x - \sin^2 x = \cos2x$$, we obtain

$$\sqrt3\sin^2x - \sin 2x = \sqrt3\cos^2x\\ \sqrt3(\cos^2x-\sin^2x) = -\sin2x\\ \sqrt3\cos2x=-\sin2x\\ \tan2x=-\sqrt3\\ 2x=-60^\circ+180^\circ k, \ k\in\mathbb{Z}$$

Whereas $$0^\circ \leqslant x \leqslant 180^\circ$$$$k=1$$ and $$120^\circ$$.

###### 10.8

Given the equation $$\sin x + \cos x = \frac1{\sqrt2}$$.

Solve it for $$-180^\circ \leqslant x \leqslant 0^\circ$$. Give your answer in degrees.

$$\sin x + \cos x = \sqrt2(\frac1{\sqrt2} \sin x + \frac1{\sqrt2} \cos x) = \sqrt2(\cos45^\circ \sin x + \sin 45^\circ \cos x) = \sqrt2 \sin(x+ 45^\circ)$$

Therefore,

$$\sqrt2 \sin(x+ 45^\circ) = \frac1{\sqrt2}\\ \sin(x+ 45^\circ) = \frac12\\ x+ 45^\circ = 30^\circ + 180^\circ k,\ k\in \mathbb Z\\ x=-15^\circ + 180^\circ k$$

Since $$-180^\circ \leqslant x \leqslant 0^\circ$$$$k =0$$ and $$x=-15^\circ$$.

###### 10.9

It is given that $$(\tan x - 1)(\sin^2x-3\cos^2x) = 0$$

Choose all possible values of $$\tan x$$.

If $$(\tan x - 1)(\sin^2x-3\cos^2x) = 0$$, then $$\tan x -1 =0$$ or $$\sin^2x =3\cos ^ 2x$$. The first equation giving that $$\tan x =1$$ and the second one giving that

$$\sin^2 x =3\cos^2 x\\ \!\tan^2 x = 3\\ \!\tan x = \pm\sqrt 3$$

###### 10.10

Given that $$\frac{\cos^2 \beta+4\sin^2 \beta}{1-\sin^2\beta} = 7$$.

Find $$\tan^2 x$$.

Since $$\sin^2 x + \cos^2 x =1$$,

$$\frac{\cos^2 \beta+4\sin^2 \beta}{1-\sin^2\beta} = 7\\ \frac{\cos^2 \beta+4\sin^2 \beta}{\cos^2\beta} = 7\\ 1+4\tan^2\beta = 7\\ 4\tan^2\beta = 6\\ \tan^2\beta = 1.5$$