Question

How do you solve for the exact solutions in the interval [0,2pi] of `sin 2x=sinx`?

Answer 1

`x = 0 , pi/3 ,(5pi)/3 ,2pi`

Explanation:

Using appropriate `color(blue)" Double angle formula "`

• sin2A = 2sinAcosA

hence : 2sinxcosx = sinx

and 2sinxcosx - sinx = 0

Take out common factor of sinx

thus : sinx(2cosx - 1 ) = 0

and so sinx = 0 or cosx =`1/2`

sinx = 0 `rArr x = 0 or 2pi`

cosx`= 1/2 rArr x = pi/3 or (2pi - pi/3 ) = (5pi)/3`

These are the solutions in the interval [0 , `2pi]`