Solve for x : sin3x=2sin^3x

sin3x = 2sin^3 x

First we
will rewrite:

sin3x =
sin(2x+x)

But we know that
:

sin(A+B) = sinAcosB +
cosAsinB

==> sin(2x+x) = sin2xcosx +
sinxcos2x

But sin2x =
2sinxcosx

==> sin3x = 2sinxcosxcosx + sinx (1-2sin^2
x)

                = 2sinxcos^2x + sinx  ( 1- 2sin^2
x)

                =2sinx ( 1-sin^2 x) + sinx ( 1- 2sin^2
x)

                 = 2sinx -2sin^3 x + sinx - 2sin^3
x

==> sin3x = 3sinx - 4sin^3 x = 2sin^3
x

    ==> 3sinx = 6sin^3
x

Divide by 3

==> sinx
= 2sin^3 x

==> 2sin^3 x - sinx =
0

==> sinx( 2sin^2 x -1) =
0

==> sinx = 0 ==> x = o, pi,
2pi

==> sin^2 x -1 =
0

==> sin^2 x =
1/2

==> sinx =
1/sqrt2

==> x = pi/4 ,
3pi/4

==> x = { 0, pi/4, 3pi/4, 2pi }
+ 2npi  n= 0, 1, 2, ....