How do you solve for 3sin^2=cos^2x?

Question

How do you solve for 3sin^2=cos^2x?

Answer 1

3sin^2=cos^2x

3sin^2 x - cos^2 x = 0

but,

sin^2 x + cos^2 x = 1

cos^2 x = 1 - sin^2 x

3sin^2 x - (1 - sin^2 x) = 0

3sin^2 x - 1 + sin^2 x = 0

4sin^2 x =  1

sin^2 x =  1/4

sin x = sqrt(1/4)

sin x = 1/2

x = arcsin (1/2)

x =30 degrees

Answer 2

3 sin ² x - cos ² x = 0

3 sin ² x - (1 - sin ² x) = 0

4 sin ² x - 1 = 0

sin ² x = 1 / 4

sin x = ± 1/2

x = 30° , 150° , 210° , 330°

Answer 3

3sin² x = cos² x

=> tan² x = 1/3

=> tan x = ± 1/√3 = tan( ± π/6 )

=> x = kπ ± π/6,   ( k ∈ Z).

Answer 4

3sin^2=1-sin^2

4sin^2=1

sin^2=1/4

sin=sqrt(1/4)

x=arcsin(sqrt(1/4))

x=pi/6 or 30degrees

make it a good day