How to solve this ?

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1 Answer
Dec 6, 2017

First, we need to know that #sin(2x)=2sin(x)cos(x)#

#sin(2x)-sin(x)=0#
#2sin(x)cos(x)-sin(x)=0#
#(sinx)(2cosx-1)=0#

we have #sinx=0 or cosx=1/2#

When #sinx=0#, #x=0 or x=pi#
When #cosx=1/2#, #x=pi/3 or x=(5pi)/3#

Therefore, the answer is not only #0 and pi#, but also include #pi/3 and (5pi)/3#.

Here is the answer :) Hope it can help you.