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

(C) Solving Trigonometric Equation (Form Quadratic Equation in sinx/ cosx/ tanx/ cosecx/ secx/ cotx)

Example:
Find all the angles between 0° and 360° that satisfy each of the following equations.
(a) 3 sin² x – 2 sin x – 1 = 0
(b)  2 sin x = cosec x + 1
(c) 5 sin² x = 2 (1 + cos x)
(d) 2 sec x = 1 + cos x
(e)  2 cot² x + 8 = 7 cosec x
Solution:
(a) 
3 sin² x – 2 sin x – 1 = 0
(3 sin x + 1)(sin x – 1) = 0
sin x = –⅓, sin x = 1
sin x = –⅓
basic angle = 19.47°
x = 180° + 19.47°, 360° – 19.47°
x = 199.47°, 340.53
sin x = 1, x = 90°
Hence x = 90°, 199.47°, 340.53°
$2\sin x=\frac{1}{\sin x}+1$

$\begin{array}{l}LHS\\ =\frac{1+\cos 2x}{\sin 2x}\\ =\frac{1+\left(2{\cos }^{2}x-1\right)}{2\sin x\cos x}\\ =\frac{\cancel{2}{\cos }^{\cancel{2}}x}{\cancel{2}\sin x\cancel{\cos x}}\\ =\frac{\cos x}{\sin x}\\ =\cot x=RHS\text{(proven)}\end{array}$


(b)
$\begin{array}{l}RHS\\ =\cot A+\tan 2A\\ =\frac{\cos A}{\sin A}+\frac{\sin 2A}{\cos 2A}\\ =\frac{\cos A\cos 2A+\sin A\sin 2A}{\sin A\cos 2A}\\ =\frac{\cos A\left({\cos }^{2}A-{\sin }^{2}A\right)+\sin A\left(2\sin A\cos A\right)}{\sin A\cos 2A}\\ =\frac{{\cos }^{3}A-\cos A{\sin }^{2}A+2{\sin }^{2}A\cos A}{\sin A\cos 2A}\\ =\frac{{\cos }^{3}A+\cos A{\sin }^{2}A}{\sin A\cos 2A}\\ =\frac{\cos A\left({\cos }^{2}A+{\sin }^{2}A\right)}{\sin A\cos 2A}\\ =\frac{\cos A}{\sin A\cos 2A}\leftarrow \boxed{{\sin }^{2}A+{\cos }^{2}A=1}\\ =\left(\frac{\cos A}{\sin A}\right)\left(\frac{1}{\cos 2A}\right)\\ =\cot A\sec 2A\end{array}$


(c)
$\begin{array}{l}LHS\\ =\frac{sinx}{1-cosx}\\ =\frac{2sin\frac{x}{2}\cos \frac{x}{2}}{1-\left(1-2si{n}^2\frac{x}{2}\right)}\leftarrow \boxed{\begin{array}{l}\sin x=2sin\frac{x}{2}\cos \frac{x}{2},\\ \cos x=1-2{\sin }^2\frac{x}{2}\end{array}}\\ =\frac{\cancel{2}\cancel{sin\frac{x}{2}}\cos \frac{x}{2}}{\cancel{2}si{n}^{\cancel{2}}\frac{x}{2}}=\frac{\cos \frac{x}{2}}{sin\frac{x}{2}}\\ =\cot \frac{x}{2}=RHS\text{(proven)}\end{array}$

(D) Solving Trigonometric Equations (Involving Addition Formulae and Double Angle Formulae)

Example 1 (Addition Formulae):
Solve the following equation for 0^o ≤ x ≤ 360^o:
$\frac{\sin x}{\cos x}=-\frac{4\sin {25}^{\circ }}{2\cos {25}^{\circ }}$
x = 137^o , 317^o

(b)

Example 2 (Double Angle):

Question 1:

Solution:
(a)(b)

$\begin{array}{l}2{\text{ sin}}^{2}x=2-\frac{x}{180}\\ 1-2{\text{ sin}}^{2}x=1-\left(2-\frac{x}{180}\right)\\ \cos 2x=\frac{x}{180}-1\\ y=\frac{x}{180}-1\\ x=0,y=-1\\ x=180,y=0\\ \\ \text{Number of solutions = 2}\end{array}$

Question 2:
(a) Sketch the graph of $y=\frac{3}{2}\cos 2x\text{ for }0\le x\le \frac{3}{2}\pi .$
(b) Hence, using the same axes, sketch a suitable straight line to find the number of solutions to the equation $\frac{4}{3\pi }x-\cos 2x=\frac{3}{2}\text{ for }0\le x\le \frac{3}{2}\pi$
State the number of solutions.

Solution:
(a)(b)
$\begin{array}{l}\frac{4}{3\pi }x-\cos 2x=\frac{3}{2}\\ \cos 2x=\frac{4}{3\pi }x-\frac{3}{2}\\ \frac{3}{2}\cos 2x=\frac{3}{2}\left(\frac{4}{3\pi }x-\frac{3}{2}\right)\\ y=\frac{2}{\pi }x-\frac{9}{4}\\ \\ \text{To sketch the graph of }y=\frac{2}{\pi }x-\frac{9}{4}\\ x=0,y=-\frac{9}{4}\\ x=\frac{3\pi }{2},y=\frac{3}{4}\\ \\ \text{Number of solutions }\\ =\text{Number of intersection points}\\ \text{= 3}\end{array}$

$\begin{array}{l}•\sin (A\pm B)=\sin A\cos B\pm \cos A\sin B\\ •\cos (A\pm B)=\cos A\cos B\mp \sin A\sin B\\ •\tan (A\pm B)=\frac{\tan A\pm \tan B}{1\mp \tan A\tan B}\end{array}$


(B) Double Angle Formulae:

(C) Half Angle Formulae:
$\boxed{\begin{array}{l}•\sin A=2\sin \frac{A}{2}\cos \frac{A}{2}\\ •\cos A={\sin }^2\frac{A}{2}-{\cos }^2\frac{A}{2}\\ \cos A=2{\cos }^2\frac{A}{2}-1\\ \cos A=1-2{\cos }^2\frac{A}{2}\\ •\tan A=\frac{2\tan \frac{A}{2}}{1-{\tan }^2\frac{A}{2}}\end{array}}$

Prove each of the following trigonometric identities.

Question 15:
$\text{Prove the identity}\frac{2}{\cos 2A+1}=se{c}^{2}A$

Solution:
$\begin{array}{l}\text{LHS}\\ \text{=}\frac{2}{\cos 2A+1}\\ =\frac{2}{\left(2{\cos }^{2}A-1\right)+1}\leftarrow \boxed{\cos 2A=2{\cos }^{2}A-1}\\ =\frac{2}{2{\cos }^{2}A}=\frac{1}{{\cos }^{2}A}\\ =se{c}^{2}A\\ =\text{RHS}\\ \therefore \text{Proven}\end{array}$
 

Question 16:
$\text{Prove the identity}\frac{2\tan A}{2-se{c}^{2}A}=\tan 2A$

Solution:
$\begin{array}{l}\text{LHS}\\ \text{=}\frac{2\tan A}{2-se{c}^{2}A}\\ =\frac{2\tan A}{2-\left({\tan }^{2}A+1\right)}\leftarrow \boxed{{\tan }^{2}A+1=se{c}^{2}A}\\ =\frac{2\tan A}{1-{\tan }^{2}A}\\ =\tan 2A\\ =\text{RHS}\\ \therefore \text{Proven}\end{array}$

Question 17:
$\text{Prove the identity}\tan x+\cot x=2\cos ec2x$

Solution:
$\begin{array}{l}\text{LHS}\\ =\tan x+\cot x\\ =\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\\ =\frac{{\sin }^{2}x+{\cos }^{2}x}{\cos x\sin x}\\ =\frac{1}{\cos x\sin x}\leftarrow \boxed{{\sin }^{2}x+{\cos }^{2}x=1}\\ =\frac{1}{\frac{1}{2}\sin 2x}\leftarrow \boxed{\begin{array}{l}\sin 2x=2\sin x\cos x\\ \frac{1}{2}\sin 2x=\sin x\cos x\end{array}}\\ =\frac{2}{\sin 2x}\\ =2\left(\frac{1}{\sin 2x}\right)\\ =2\cos ec2x\\ =\text{RHS}\\ \therefore \text{Proven}\end{array}$
 

Question 18:
$\text{Prove the identity}\frac{\cos x-\sin 2x}{\cos 2x+\sin x-1}=\frac{1}{\tan x}$

Solution:
$\begin{array}{l}\text{LHS}\\ =\frac{\cos x-\sin 2x}{\cos 2x+\sin x-1}\\ =\frac{\cos x-2\sin x\cos x}{\left(1-2{\sin }^{2}x\right)+\sin x-1}\leftarrow \boxed{\cos 2x=1-2{\sin }^{2}x}\\ =\frac{\cos x\left(1-2\sin x\right)}{\sin x-2{\sin }^{2}x}\\ =\frac{\cos x\left(1-2\sin x\right)}{\sin x\left(1-2\sin x\right)}\\ =\frac{\cos x}{\sin x}\\ =\cot x\\ =\frac{1}{\tan x}\\ =\text{RHS}\\ \therefore \text{Proven}\end{array}$

5.2a Six Trigonometric Functions of Any Angle 

(A) The definition of sin, cos, tan, cosec, sec and cot