We are aware that the trigonometric functions which are ratios of the pair of sides of the right angled triangle containing the acute angle. The trigonometric functions are periodic. When we observe the graph of sine and cosine functions we observe that the graph reaches max and minimum point and continues the same. Also the graph of tangent function likes between consecutive multiples of $\frac{\pi}{2}$. Equations are those where we find the variable and solve for the unknown variable. When we solve the equations containing the trigonometric functions we will be solving for the unknown angle of the given trigonometric function. In this section let us see how we solve equations containing trigonometric functions.

Trigonometric Equations are the equations involving trigonometric function or functions of unknown angles.

Example: sin x = 0, cos2 x - sin2 x = 0 etc.

Solution: A solution of a trigonometric equation is a value of the unknown angle that satisfies the equation.
A trigonometric equation may have unlimited number of solutions.
for example, if sin x = 0,
x = 0, $\pm$ $\pi$, $\pm$ 2 $\pi$, $\pm$ 3 $\pi$, ..........................

Principal solution: The solution lying between 0 and 2 $\pi$ ( 0o and 360o ) is called a principal solution.

General solution: Since the trigonometric functions are periodic, a solution generalized by means of periodicity is known as the general solution.

Every trigonometric equation will have principal solution as well as general solution.
We always find the general solution if not otherwise stated.

The principal solution can be obtained from the general solution.

Example:

Find the Principal solution of the equation cos x = - $\frac{1}{2}$

Solution: We know that cos $\frac{\pi}{3}$ = 1
and cos is negative in the second and third quadrant

( i. e ) cos ( $\pi$ - $\frac{\pi}{3}$ ) = -1

and cos ( $\pi$ + $\frac{\pi}{3}$ ) = -1

Therefore, we have x = ( $\pi$ - $\frac{\pi}{3}$ ) and ( $\pi$ - $\frac{\pi}{3}$ ).
(i. e ) x = 2$\frac{\pi}{2}$ and 4$\frac{\pi}{3}$ are the principal solutions.

The Equations, sin x = 0, cos x = 0 and tan x = 0

Let us consider the following equations:
1. Sin x = 0
Proof :
since sin 0 = 0
sin $\pi$ = 0
sin 2$\pi$ = 0
sin [- $\pi$ ] = 0
sin [-2$\pi$ ] = 0
.........................
we have in general,
sin n$\pi$ = 0, where n is an integer.
Therefore, the solution of the equation sin x = 0 is
x = n $\pi$, where n $\epsilon$ Z [ where Z is an integer ]


2. Cos x = 0
Proof :since cos $\frac{\pi}{2}$ = 0

cos 3$\frac{\pi}{2}$ = 0

cos 5$\frac{\pi}{2}$ = 0

cos [-$\frac{\pi}{2}$] = 0

cos [-3$\frac{\pi}{2}$] = 0
........................
we have in general,
cos n$\pi$ = 0, where n is an integer.
Therefore, the solution of the equation cos x = 0 is
x = (2n+1) $\frac{\pi}{2}$, where n $\epsilon$ Z [ where Z is an integer ]


3. tan x = 0
Proof: This equation is satisfied when x = 0, $\pm$ 1 $\pi$, $\pm$ 2 $\pi$, $\pm$ 3 $\pi$, .....................
In general x = n $\pi$ , where n $\epsilon$ Z
Let us solve the following examples:

Solved Examples

Question 1: Solve sin x = 0.5 and find its principal solution.
Solution:
 
We have sin x = 0.5
using inverse trigonometric functions, we have

     x = arcsin ( 0.5 ) = 30o = $\frac{\pi}{6}$

since sine is positive in the first and the second quadrant.

we have x = $\frac{\pi}{6}$ and $\pi$ - $\frac{\pi}{6}$ = 5$\frac{\pi}{6}$

Therefore, the principal solutions are, x = $\frac{\pi}{6}$ and 5$\frac{\pi}{6}$
 

Question 2: Solve the equation and find its general solution:
  3 sin2 x + 2 sin x = 5
Solution:
 
Let sin x = u
Therefore, the above equation is
                       3 u2 + 2 u = 5
            =>      3 u2 + 2 u - 5 = 0
Factoring the above equation by splitting the middle term,
                        3 u2 + 5 u - 3 u - 5 = 0
        =>  u ( 3 u + 5 ) - 1 ( 3 u + 5 ) = 0
       =>              ( u - 1 ) ( 3 u + 5 ) = 0
       =>                                 u - 1 = 0 ( o r ) 3 u + 5 = 0
       =>                                     u = 1 or u = - $\frac{5}{3}$

       =>                                sin x = -1 , as sin x $\neq$ $\frac{5}{3}$ [ substituting u = sin x ]

                                                x = arcsin (-1)

                                                   = 3$\frac{\pi}{2}$
Therefore, the principal solution to the above equation is 3$\frac{\pi}{2}$
 

Let us discuss with some of the trigonometric equations and their general solutions.
1. sin x = sin $\alpha$ is
x = n $\pi$ + ( -1 )n $\alpha$,

2. cos x = cos $\alpha$ is
x = 2n $\pi$ $\pm$ $\alpha$, where n $\epsilon$ Z

3. tan x = tan $\alpha$ is
x = n $\pi$ + $\alpha$, where n $\epsilon$ Z.

We use the trigonometric identities and other multiple angle formulas to simplify the trigonometric equations.

Let us discuss the following example.

Solved Example

Question: Find all the values of x satisfying the equation sin x + sin 5 x = sin 3 x, such that 0 $\le$ x $\le$ $\pi$
Solution:
 
We have sin x + sin 5 x = sin 3 x
                   =>                2 sin 3x. cos 2x = sin 3x
                   => 2 sin 3x . cos 2x - sin 3x   = 0
                   => sin 3x ( 2 cos 2x - 1 )        = 0
                    =>                             sin 3x = 0 ( o r ) 2 cos 2x - 1 = 0
First Solution:                               sin 3x = 0
                    =>                                 3 x = 0,  $\pi$, 3$\pi$

                    =>                                    x = 0, $\frac{\pi}{3}$$\frac{3\pi}{3}$

                    =>                                    x = 0 , $\frac{\pi}{3}$, $\pi$
Similarly,
                                            2 cos 2x - 1 = 0
                   =>                             cos 2x = $\frac{1}{2}$

                                                              = cos $\frac{\pi}{3}$

                                                              = cos (2 $\pi$ - $\frac{\pi}{3}$)

                  =>                                  2 x = $\frac{\pi}{3}$, $\frac{5\pi}{3}$

                  =>                                     x = $\frac{\pi}{6}$, $\frac{5\pi}{6}$

other values are rejected as they are more than $\pi$

Therefore all the possible values of x are, 0 , $\frac{\pi}{6}$, $\frac{\pi}{3}$, $\frac{2\pi}{3}$, $\frac{5\pi}{6}$, $\pi$.
 


By drawing the above equation in the graphing calculator we find that the curve intersects the x-axis at 6 points

which are 0, $\frac{\pi}{6}$, $\frac{\pi}{3}$, $\frac{2\pi}{3}$, $\frac{5\pi}{6}$, $\pi$.

Therefore, the graphical solution of the function y = sin x + sin 5 x - sin 3 x, where 0 $\le$ x $\le$ $\pi$ are x = 0 , $\frac{\pi}{6}$, $\frac{\pi}{3}$, $\frac{2\pi}{3}$, $\frac{5\pi}{6}$, $\pi$.

Solved Example

Question: Solve sin 9x = sin x, for 0 $\le$ x $\le$ 360o
Solution:
 
We have sin 9x = sin x
The solution will be,
                            sin 9x = sin x or
                             sin 9x = sin ( $\pi$ - x )

First solution: sin 9x = sin x

                            => 9x = x, 2 $\pi$ + x , 4$\pi$ + x, 6 $\pi$ + x, 8 $\pi$ + x, 10 $\pi$ + x, . . . . . . . . . .
                            => 9 x - x = 0, 2 $\pi$, 4 $\pi$. 6 $\pi$. 8 $\pi$. 10 $\pi$, . . . . . . . . . .
                           =>       8 x = 0, 2 $\pi$, 4 $\pi$. 6 $\pi$. 8 $\pi$. 10 $\pi$, . . . . . . . . . .

                           =>      8  x = 0, $\frac{2\pi}{2}$, $\frac{4\pi}{2}$, $\frac{6\pi}{2}$, $\frac{8\pi}{2}$, $\frac{10\pi}{2}$, $\frac{12\pi}{2}$, , , , , , ,

                           =>      8  x = 0 , $\pi$, 2 $\pi$, 4 $\pi$. 6 $\pi$. 8 $\pi$. 10 $\pi$,  , , , , , ,

                                         x = 0, $\frac{2\pi}{8}$, $\frac{4\pi}{8}$, $\frac{6\pi}{8}$, $\frac{8\pi}{8}$, , , , , ,

                                         x   = 0, $\frac{\pi}{4}$, $\frac{\pi}{2}$, $\frac{3\pi}{4}$, $\pi$, $\frac{5\pi}{4}$, $\frac{3\pi}{2}$, $\frac{7\pi}{2}$,  2 $\pi$.        

Second Solution: sin 9x = sin ( $\pi$ - x )

                                        9x =  ( $\pi$ - x ), ( 3 $\pi$ - x ), ( 5 $\pi$ - x ), (7  $\pi$ - x ), ( 9 $\pi$ - x ), ( 11 $\pi$ - x ),
                                                                                                       ( 13  $\pi$ - x ), ( 15 $\pi$ - x ), ( 17 $\pi$ - x ) (19  $\pi$ - x ).
                            =>      10 x = $\pi$, 3 $\pi$, 5 $\pi$, 7 $\pi$, 9 $\pi$, 11 $\pi$, 13 $\pi$, 15 $\pi$, 17 $\pi$, 19 $\pi$
                            =>          x = $\frac{\pi}{10}$, $\frac{3\pi}{10}$, $\frac{5\pi}{10}$, $\frac{7\pi}{10}$, $\frac{9\pi}{10}$, , , , , , , , ,, $\frac{19\pi}{10}$