Solve the equation, X is real number. 2cos(6x) + 4sin(3x) = 3

2cos(6x) + 3sin(3x)= 3

First,
we know that:

cos2x = 2sin^2 x -
1

==> cos6x = 1-2sin^2
3x

Now we will substitute into the
equation.

==> 2(1-2sin^2 3x) + 4sin3x =
3

==> 2 - 4sin^2 3x + 4sin3x -3 =
0

==> -4sin62 3x + 4sin3x -1 =
0

Multiply by -1:

==>
4sin^2 3x - 4sin3x +1 = 0

Now we will
factor:

==> (2sin3x -1)^2 =
0

==> (2sin3x -1)=
0

==> sin3x =
1/2

==> 3x = pi/6+2npi , 5pi/6+2npi  (n= 0, 1, 2,
3...)

==> x = pi/18 ,
5pi/18

==> x = { pi/18+2npi ,
5pi/18+2npi}