How do you solve for x. 3sin^(2)x = cos^(2)x ; 0 less than or equal to x which is less than or equal to 2pi?

Question

Please show steps because i have more of them and i want to learn how to do it.

Answer 1

3 sin ² x = 3 - 3 cos² x= cos² x ---> cos² x = 3/4--> cos x = ± √3 / 2 ---> x = π / 6 , 5π / 6 , 7π / 6 , 11π / 6

Answer 2

Use the pythagorean relation: sin^2(x) = 1 - cos^2(x)

3*sin^2(x) = 3 * (1 - cos^2(x)) = 3 - 3*cos^2(x)

cos^2(x) = 3 - 3*cos^2(x)

4*cos^2(x) = 3

cos(x) = sqrt(3)/2

x = pi / 6

cos^2(x) and sin^2(x) are always positive, so you have valid answers in all four quadrants.

x = pi/6 (30 deg), 5 pi / 6 (150 deg), 7 pi / 6 (210 deg) and 11 pi / 6 (330 deg).