solve 3cos^2 2x +4 sin 2x =1
0<x<360
3cos^2 (2x) + 4sin(2x) - 1 = 0
3(1 - sin^2 (2x) ) + 4sin(2x) - 1 = 0
I will let Ø = 2x for easier typing
3 - 3sin^2 Ø + 4sinØ - 1 = 0
3sin^2 Ø - 4sinØ - 2 = 0
does not factor, too bad, so
sinx = (4 ± √40)/6 = -.38742... or -1.72..
but the sine of anything has to be between -1 and +1
so sinØ = -.38742..
Ø must be in III or IV
Ø = 180+22.7944 OR Ø = 360 - 22.7944
Ø = 202.79.. or Ø = 337.205...
x = 101.4° or x = 168.6°
but the period of both cos 2x and sin 2x is 180°
so adding or subtracting 180 from any answer gets a new one
we could also have
x = 101.4 + 180 = 281.4°
x = 168.6 + 180 = 348.6°
x = 101.4° , 168.6° , 281.4° , 348.6°