# Question #bb853

Oct 21, 2016

1(a)

given $\tan 2 \theta = - \frac{20}{21}$
Let $\tan \theta = x$

So using the identity $\tan 2 \theta = \frac{2 \tan \theta}{1 - {\tan}^{2} \theta}$ we get

$- \frac{20}{21} = \frac{2 x}{1 - {x}^{2}}$

$\implies - \frac{10}{21} = \frac{x}{1 - {x}^{2}}$

$\implies 10 {x}^{2} - 21 x - 10 = 0$

$\implies 10 {x}^{2} - 25 x + 4 x - 10 = 0$

$\implies 5 x \left(2 x - 5\right) + 2 \left(2 x - 5\right) = 0$

$\implies \left(2 x - 5\right) \left(5 x + 2\right) = 0$

So $x = \frac{5}{2} \mathmr{and} x = - \frac{2}{5}$

Hence $\tan \theta = \frac{5}{2} \mathmr{and} \tan \theta = - \frac{2}{5}$
$1(b) given $\tan 2 \theta = - \frac{36}{77}$ Let $\tan \theta = x$ So using the identity $\tan 2 \theta = \frac{2 \tan \theta}{1 - {\tan}^{2} \theta}$ we get $\frac{36}{77} = \frac{2 x}{1 - {x}^{2}}$ $\implies \frac{18}{77} = \frac{x}{1 - {x}^{2}}$ $\implies 18 {x}^{2} + 77 x - 18 = 0$ $\implies 18 {x}^{2} + 81 x - 4 x - 18 = 0$ $\implies 9 x \left(2 x + 9\right) - 2 \left(2 x + 9\right) = 0$ $\implies \left(2 x + 9\right) \left(9 x - 2\right) = 0$ So $x = - \frac{9}{2} \mathmr{and} x = \frac{2}{9}$ Hence $\tan \theta = - \frac{9}{2} \mathmr{and} \tan \theta = \frac{2}{9}$$
2(a)
$\cos 2 \theta + 5 \cos \theta = 2$

$\implies 2 {\cos}^{2} \theta - 1 + 5 \cos \theta - 2 = 0$

$\implies 2 {\cos}^{2} \theta + 5 \cos \theta - 3 = 0$

$\implies 2 {\cos}^{2} \theta + 6 \cos \theta - \cos \theta - 3 = 0$

$\implies \left(2 \cos \theta - 1\right) \left(\cos \theta + 3\right) = 0$

So $\cos \theta = - 3 \to \text{not possible}$

when
$\left(2 \cos \theta - 1\right) = 0$

$\implies \cos \theta = \frac{1}{2} = \cos {60}^{\circ} = \cos {\left(360 - 60\right)}^{\circ} = \cos {300}^{\circ}$

Hence $\theta = {60}^{\circ} \mathmr{and} \theta = {300}^{\circ}$

$2(b) $2 \sec 2 x - \cot 2 x = \tan 2 x$ $\implies 2 \sec 2 x = \tan 2 x + \cot 2 x$ $\implies 2 \sec 2 x = \frac{\sin 2 x}{\cos 2 x} + \frac{\cos 2 x}{\sin 2 x}$ $\implies 2 \sec 2 x = \frac{{\sin}^{2} 2 x + {\cos}^{2} 2 x}{\sin 2 x \cos 2 x} = \csc 2 x \sec 2 x$ $\implies 2 \sec 2 x - \csc 2 x \sec 2 x = 0$ $\implies \sec 2 x \left(2 - \csc 2 x\right) = 0$ $\sec 2 x = 0 \to \text{not possible}$ when $\left(2 - \csc 2 x\right) = 0$ $\sin 2 x = \frac{1}{2} = \sin {30}^{\circ} = \sin \left(180 - 30\right) = \sin {150}^{\circ}$ Hence $x = {15}^{\circ} \mathmr{and} x = {75}^{\circ}$$`
2(c)

$2 \tan \left(\frac{\theta}{2}\right) + 3 \tan \theta = 0$

$\implies 2 \tan \left(\frac{\theta}{2}\right) + \frac{6 \tan \left(\frac{\theta}{2}\right)}{1 - {\tan}^{2} \left(\frac{\theta}{2}\right)} = 0$

$\implies 2 \tan \left(\frac{\theta}{2}\right) \left(1 + \frac{3}{1 - {\tan}^{2} \left(\frac{\theta}{2}\right)}\right) = 0$

$\implies 2 \tan \left(\frac{\theta}{2}\right) \left(4 - {\tan}^{2} \left(\frac{\theta}{2}\right)\right) = 0$
when
$\tan \left(\frac{\theta}{2}\right) = 0 = \tan \left({o}^{\circ}\right) = \tan {180}^{\circ}$

Hence $\theta = {0}^{\circ} \mathmr{and} \theta = {360}^{\circ}$

$\left(4 - {\tan}^{2} \left(\frac{\theta}{2}\right)\right) = 0$

$\implies \tan \left(\frac{\theta}{2}\right) = \pm 2$
when
$\implies \tan \left(\frac{\theta}{2}\right) = + 2 = \tan {63.43}^{\circ}$

$\implies \theta = {126.86}^{\circ}$

when
$\implies \tan \left(\frac{\theta}{2}\right) = - 2 = - \tan {63.43}^{\circ} = \tan \left(180 - 63.43\right)$

$\implies \tan \left(\frac{\theta}{2}\right) = \tan {116.57}^{\circ}$

$\implies \theta = {233.14}^{\circ}$