Answer:
Consider the given equation.
4sinxsin2xsin4x=sin3x
2(2sin2xsinx)sin4x=sin3x
We know that
2sinAsinB=cos(A−B)−cos(A+B)
Therefore,
2[cos(2x−x)−cos(2x+x)]sin4x=sin3x
2[cosx−cos3x]sin4x=sin3x
2sin4xcosx−2sin4xcos3x=sin3x
We know that
2sinAcosB=sin(A+B)+sin(A−B)
Therefore,
sin(4x+x)+sin(4x−x)−[sin(4x+3x)+sin(4x−3x)]=sin3x
sin(5x)+sin(3x)−[sin(7x)+sin(x)]=sin3x
sin(5x)−sin(7x)=sin(x)
We know that
sinC−sinD=2cos(
2
C+D
)sin(
2
C−D
)
Therefore,
2cos(
2
5x+7x
)sin(
2
5x−7x
)=sinx
2cos(
2
12x
)sin(
2
−2x
)=sinx
2cos(6x)sin(−x)=sinx
−2cos(6x)sin(x)=sinx
cos6x=−
2
1
6x=2nπ±
3
2π
x=
3
nπ
±
9
π
Hence, the value is
3
nπ
±
9
π
.
