Trig equation

A quick graph shows pro35hp is clearly correct:

Subtract 12 from each side

will get $\sin \left(2x\right)+2\left(2\cos \left(x\right)-3\sin \left(x\right)\right)-12=0$

Use the identity $\sin \left(2x\right)=2\cos \left(x\right)\sin \left(x\right)$

$\left(2\cos \left(x\right)-3\sin \left(x\right)\right)2+2\cos \left(x\right)\sin \left(x\right)-12=0$

$2\left(2\cos \left(x\right)+\cos \left(x\right)\sin \left(x\right)-3\sin \left(x\right)-6\right)$

$=2\left(\sin \left(x\right)+2\right)\left(\cos \left(x\right)-3\right)$

solving each part equivalent to 0

$\sin \left(x\right)+2=0\:\:\:\mathrm{or}\:\:\:\cos \left(x\right)-3=0$

sinx cannot be -2

and cos x cannot be 3 

hence , no answer in R

Solve for x:
2 (2 cos(x)-3 sin(x))+sin(2 x) = 12

2 (2 cos(x)-3 sin(x))+sin(2 x)  =  4 cos(x)-6 sin(x)+sin(2 x):
4 cos(x)-6 sin(x)+sin(2 x) = 12

Subtract 12 from both sides:
-12+4 cos(x)-6 sin(x)+sin(2 x) = 0

Simplify trigonometric functions:
2 (cos(x)-3) (2+sin(x)) = 0

Divide both sides by 2:
(cos(x)-3) (2+sin(x)) = 0

Split into two equations:
cos(x)-3 = 0 or 2+sin(x) = 0

Add 3 to both sides:
cos(x) = 3 or 2+sin(x) = 0

Take the inverse cosine of both sides:
x = cos^(-1)(3)+2 pi n_1  for  n_1  element Z or x = 2 pi n_2-cos^(-1)(3)  for  n_2  element Z
   or  2+sin(x) = 0

Subtract 2 from both sides:
x = cos^(-1)(3)+2 pi n_1  for  n_1  element Z
   or  x = 2 pi n_2-cos^(-1)(3)  for  n_2  element Z
   or  sin(x) = -2

Take the inverse sine of both sides:
Answer: |  x = cos^(-1)(3)+2 pi n_1  for  n_1  element Z
   or  x = 2 pi n_2-cos^(-1)(3)  for  n_2  element Z
   or  x = pi+sin^(-1)(2)+2 pi n_3  for  n_3  element Z                                                                           or x = 2 pi n_4- sin^(-1)(2)  for  n_4  element Z