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 \(\sin^2 x + \cos^2 x =1\),

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