5.6b Solving Trigonometric Equation (Factorization)

(b)  $3\sec x=4\cos x$  
(c)  $16\tan x=\cot x$

Solution:
(a)
$\begin{array}{l}\cot x=-2\cos x\\ \frac{\cos x}{\sin x}=-2\cos x\\ \cos x=-2\cos x\sin x\\ \cos x+2\sin x\cos x=0\\ \cos x\left(1+2\sin x\right)=0\\ \cos x=0\\ x={90}^{\circ },{270}^{\circ }\\ 1+2\sin x=0\\ \sin x=-\frac{1}{2}\\ \text{Basic}\angle =\text{3}{0}^{\circ }\\ x=\left({180}^{\circ }+{30}^{\circ }\right),\left({360}^{\circ }-{30}^{\circ }\right)\\ x={210}^{\circ },{330}^{\circ }\\ \therefore x={90}^{\circ },{210}^{\circ },{270}^{\circ },{330}^{\circ }\end{array}$

(b)
$\begin{array}{l}3\sec x=4\cos x\\ \frac{3}{\cos x}=4\cos x\\ 3=4{\cos }^{2}x\\ {\cos }^{2}x=\frac{3}{4}\\ \cos x=\pm \frac{\sqrt{3}}{2}\\ \text{Basic}\angle ={30}^{\circ }\\ x={30}^{\circ },\left({180}^{\circ }-{30}^{\circ }\right),\left({180}^{\circ }+{30}^{\circ }\right),\left({360}^{\circ }-{30}^{\circ }\right)\\ x={30}^{\circ },{150}^{\circ },{210}^{\circ },{330}^{\circ }\end{array}$

(c)
$\begin{array}{l}16\tan x=\cot x\\ 16\tan x=\frac{1}{\tan x}\\ {\tan }^{2}x=\frac{1}{16}\\ \tan x=\pm \frac{1}{4}\\ \text{Basic}\angle ={14.04}^{\circ }\\ x={14.04}^{\circ },\left({180}^{\circ }-{14.04}^{\circ }\right),\\ \left({180}^{\circ }+{14.04}^{\circ }\right),\left({360}^{\circ }-{14.04}^{\circ }\right)\\ x={14.04}^{\circ },{165.96}^{\circ },{194.04}^{\circ },{345.96}^{\circ }\end{array}$