# How many roots in equation sinx=sin2x in 0,2pie?

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You need to solve the trigonometric equation sin x = sin 2x.

Notice that moving sin 2x to the left side yields a difference of two like functions, hence you may use transformation of like functions in products such that:

`sin x - sin 2x = 2 (cos (x+2x)/2)*(sin (x - 2x)/2)`

`sin x - sin 2x = 2 (cos (3x)/2)* (sin(-x)/2)`

Use the fact that sine function is odd such that:

`sin x - sin 2x = -2 (cos (3x)/2)* (sin(x)/2)`

If `sin x - sin 2x = 0 =gt -2 (cos (3x)/2)* (sin(x)/2) = 0`

`cos (3x)/2 = 0 =gt (3x)/2 = cos^(-1) 0`

`` `3x/2 = pi/2 =gt x = pi/3`

You need to remember that values of cosine function are positive in the fourth quadrant, hence x = `2pi - pi/3` => x = `5pi/3` .

sin `x/2` = 0 => `x/2` = 0 => x = 0

**The problem does no tell if the values 0 or 2pi are accepted, hence, considering these values accepted, the number of solutions to this equation is 3: `x = 0, x = pi/3, x = 5pi/3` .**