(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:
$\sin A=\frac{2}{3}\leftarrow \left(\text{sin A is positive in the 1st and 2nd quadrants}\text{.}\right)$

Question 2:

Solution:
cos x =  $-\frac{3}{4}$

Question 3:

Solution:
6 (1 + tan²x) – 13 tan x = 0
6 tan²x – 13 tan x + 6 = 0
(3 tan x – 2)(2 tan x – 3) = 0
tan x = 2/3   or   tan x = 3/2

tan x = 2/3
Basic angle = 33.69°
x = 33.69°, 180° + 33.69°
x = 33.69°, 213.69° 
Or
tan x = 3/2
Basic angle = 56.31°
x = 56.31°, 180° + 56.31°
x = 56.31°, 236.31°

Hence x = 33.69°, 56.31°, 213.69°, 236.31°. 

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


(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

5.6 Simple Trigonometric Equations

Steps to solve simple trigonometric equation:
(1) Determine the range of values of the required angles.
(2) Find a basic angle by using calculator.
(3) Determine the quadrants the angle should be.
(4) Determine the values of angles in those quadrants.

(A) Solving Trigonometric Equation (Basic Equation in sinx/ cosx/ tanx/ cosecx/ secx/ cotx)

Example:
Find all the values of θ  for 0° < θ  < 360° that satisfy each of the following trigonometric equations.
(a) sin θ = 0.6137
(b) cos θ = 0.2377
(c) tan θ = 2.7825
(d) sin θ = -0.8537

Solution:

Question 9:

Solution:
(a)


$\sin \theta =\frac{3}{5}\cos \theta =\frac{4}{5}\tan \theta =\frac{3}{4}$


(c)
$\begin{array}{l}\tan \left({360}^{\circ }+\theta \right)\\ \\ =\frac{\tan {360}^{\circ }+\tan \theta }{1-\tan {360}^{\circ }\tan \theta }\\ \\ =\frac{0+\tan \theta }{1-\left(0\right)\left(\tan \theta \right)}\\ =\tan \theta \\ =\frac{3}{4}\end{array}$

$\begin{array}{l}\text{LHS:}{\cot }^{2}x-{\cot }^{2}x{\cos }^{2}x\\ ={\cot }^{2}x\left(1-{\cos }^{2}x\right)\\ ={\cot }^{2}x\left(si{n}^{2}x\right)\\ =\frac{{\cos }^{2}x}{si{n}^{2}x}\left(si{n}^{2}x\right)\\ ={\cos }^{2}x\text{(RHS)}\end{array}$


(b)
$\begin{array}{l}\text{LHS:}\frac{\sec x}{\sec x-\cos x}\\ =\frac{\frac{1}{\cos x}}{\frac{1}{\cos x}-\cos x}=\frac{\frac{1}{\cos x}}{\frac{1}{\cos x}-\frac{{\cos }^{2}x}{\cos x}}\\ =\frac{\frac{1}{\cos x}}{\frac{1-{\cos }^{2}x}{\cos x}}=\frac{1}{\cos x}\times \frac{\cos x}{1-{\cos }^{2}x}\\ =\frac{1}{1-{\cos }^{2}x}=\frac{1}{si{n}^{2}x}\\ =\cos e{c}^{2}x\text{(RHS)}\end{array}$