# How to solve the equation sin 3x=sin x, if 0<x<pi, and not using triple angle formula?

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### 2 Answers

We have to solve sin 3x = sin x for 0 < x < pi.

Use sin(A + B) = sin A cos B + cos A sin B

sin 3x = sin (2x + x)

=> sin 2x * cos x + cos 2x * sin x

=> (sin x * cos x + sin x* cos x) cos x + (cos x * cos x - sin x * sin x)*sin x

=> 2* sin x * (cos x)^2 + (cos x)^2*sin x - (sin x)^3

=> 3*sin x * (cos x)^2 - (sin x)^3

3*sin x * (cos x)^2 - (sin x)^3 = sin x

=> 3*(cos x)^2 - (sin x)^2 = 1

=> 3 - 3*(sin x)^2 - (sin x)^2 = 1

=> 3 - 4*(sin x)^2 = 1

=> 4*(sin x)^2 = 2

=> (sin x)^2 = 1/2

=> sin x= 1/sqrt 2

x = arc sin (1/sqrt 2)

=> x = pi/4

**The required solution is x = pi/4**

Well, we can transform the difference that we'll get if we'll move the term sin x to the left side, into a product:

sin 3x - sin x = 0

2[cos (3x+x)/2]*[sin(3x - x)/2] = 0

2 cos 2x*sin x = 0

We'll divide by 2:

cos 2x*sin x = 0

We'll set each factor as zero:

cos 2x = 0

2x = arccos 0

2x = pi/2

x = pi/4

sin x = 0

x = arcsin 0

x = 0 or pi

**The value of x, found in the range 0<x<pi = pi/4.**