How do you solve Cos^2x + 3*Sin^2x + sqrt3*Sin2x = 1

How do you solve Cos^2x + 3*Sin^2x + sqrt3*Sin2x = 1

\(Cos^2x + 3Sin^2x + \sqrt3*Sin2x = 1\\ Cos^2x +Sin^2x+ 2Sin^2x + \sqrt3*Sin2x = 1\\ 1+ 2Sin^2x + \sqrt3*Sin2x = 1\\ 2Sin^2x + \sqrt3*Sin2x = 0\\ 2Sin^2x + \sqrt3*2sinxcosx = 0\\ 2Sinx(sinx + \sqrt3*cosx )= 0\\ sinx=0 \qquad or \qquad sinx + \sqrt3*cosx=0\\ sinx=0 \qquad or \qquad sinx =- \sqrt3*cosx\\ sinx=0 \qquad or \qquad tanx =- \sqrt3\\ x=0,\;\; 180^0,\;\; 120^0,\;\;300^0\\ x=180n \qquad or\quad x=180n-60\qquad n\in Z\\ \mbox{All angles measurements are in degrees} \)

Solve for x:
cos^2(x)+3 sin^2(x)+sqrt(3) sin(2 x) = 1

Simplify trigonometric functions:
2-2 sin(pi/6-2 x) = 1

Subtract 2 from both sides:
-2 sin(pi/6-2 x) = -1

Divide both sides by -2:
sin(pi/6-2 x) = 1/2

Take the inverse sine of both sides:
pi/6-2 x = (5 pi)/6+2 pi n_1  for  n_1  element Z
   or  pi/6-2 x = pi/6+2 pi n_2  for  n_2  element Z

Subtract pi/6 from both sides:
-2 x = (2 pi)/3+2 pi n_1  for  n_1  element Z
   or  pi/6-2 x = pi/6+2 pi n_2  for  n_2  element Z

Divide both sides by -2:
x = -pi/3-pi n_1  for  n_1  element Z
   or  pi/6-2 x = pi/6+2 pi n_2  for  n_2  element Z

Subtract pi/6 from both sides:
x = -pi/3-pi n_1  for  n_1  element Z
   or  -2 x = 2 pi n_2  for  n_2  element Z

Divide both sides by -2:
Answer: |  x = -pi/3-pi n_1  for  n_1  element Z
            or  x = -pi n_2  for  n_2  element Z

How do you solve Cos^2x + 3*Sin^2x + sqrt3*Sin^2x = 1  ?

Thanks Omi :)

Omi thinks that you may have forgotten a ^ symbol.

She has answered a possible alternate version of your question.

I made a small mistake. It has to be called: