What is the general solution for 3sin^2(θ)+2cos(2θ)=1 ?

The general solution for 3sin^2(θ)+2cos(2θ)=1 is θ= pi/2+2pin,θ=(3pi)/2+2pin

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3sin^2(θ)+2cos(2θ)=1

3 sin^{2} (θ) + 2 cos (2 θ) = 1

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Solution steps

3 sin^{2} (θ) + 2 cos (2 θ) = 1

Subtract

1

from both sides

3 sin^{2} (θ) + 2 cos (2 θ) - 1 = 0

Rewrite using trig identities

1 - sin^{2} (θ) = 0

Solve by substitution

sin (θ) = 1, sin (θ) = - 1

sin (θ) = 1 : θ = \frac{π}{2} + 2 πn

sin (θ) = - 1 : θ = \frac{3 π}{2} + 2 πn

Combine all the solutions

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

sin^{4} (x) = 1

2 sin (x) cos (x) + 2 sin (x) = 0, 0 \leq x < 2 π

(cos (x) + 1) = 0

cos (x) = - 1, sin (x)

3 cos (2 θ) - 5 cos (θ) = 1

What is the general solution for 3sin^2(θ)+2cos(2θ)=1 ?
The general solution for 3sin^2(θ)+2cos(2θ)=1 is θ= pi/2+2pin,θ=(3pi)/2+2pin