Trig Q - find all values of x between 0 < x < 360 satisfying..

5cos² + 2 = 3sin²x - 2cos²

Please help

5cos^2x + 2 = 3sin^2x - 2cos^2x
7cos^2x + 2 = 3(1 - cos^2x)
7cos^2x + 2 = 3 - 3cos^2x
10cos^2x = 1
cos^2x = 1/10
cosx = 1/√10 or cosx = -1/√10
x = 71.6 or x = 288.4 or x = 108.4 or x = 251.6

7cos^2 (x) - 3sin^2 (x) + 2 = 0
7(1 - sin^2(x)) - 3sin^2(x) + 2 = 0
7 - 7sin^2(x) - 3sin^2(x) + 2 = 0
7 - 10sin^2(x) + 2 = 0
9 - 10sin^2(x) = 0
10sin^2(x) = 9
sin^2(x) = 9/10
sin(x) = sqrt(9/10) = 3/sqrt(10)

Could use trig to find one of the angles.
Using a graphing calculator, the angle is 71.56 degrees.

7cos^2x-3sin^2x+2=0
7(1-sin^2x)-3sin^2x+2=0
7-7sin^2x-3sin^2x+2=0
9-10sin^2x= 0
10sin^2x=9
sin^2x= 9/10
sinx= +- sqrt(9/10)
x= 71.56 degrees, 180-71.56= 108.44 degrees, 71.56+180= 251.56 degrees
and 360-71.56= 288.44 degrees.

7cos²x + 2 = 3(1 - cos²x)
7cos²x + 2 = 3 - 3cos²x
10cos²x = 3 - 2
10cos²x = 1
cos²x = 1/10