sinx+sin5x=sin3x Using sin2A+sin2B=2sin(A+B)cos(A−B), we get: 2sin3xcos2x=sin3x ⇒sin3x(2cos2x−1)=0 ⇒sin3x=0 OR 2cos2x−1=0 CASE 1:sin3x=0=sinnπ,n∈Z. ⇒x=
nπ
3
⇒x=...,0,
π
3
,
2π
3
,π,... CASE 2:2cos2x−1=0 ⇒cos2x=
1
2
=cos(2nπ±
π
3
),n∈Z. ⇒x=nπ±
π
6
=(6n±1)
π
6
⇒x=...,
π
6
,
5π
6
,... Therefore, there are 6 possible values of x in the interval [0,π]