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

The general solution for 2sin^2(θ)-cos(2θ)=1 is θ= pi/4+pin,θ=(3pi)/4+pin

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2sin^2(θ)-cos(2θ)=1

2 sin^{2} (θ) - cos (2 θ) = 1

θ = \frac{π}{4} + πn, θ = \frac{3 π}{4} + πn

Solution steps

2 sin^{2} (θ) - cos (2 θ) = 1

Subtract

1

from both sides

2 sin^{2} (θ) - cos (2 θ) - 1 = 0

Rewrite using trig identities

- 2 cos (2 θ) = 0

Divide both sides by

- 2

cos (2 θ) = 0

General solutions for

cos (2 θ) = 0

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

Solve

2 θ = \frac{π}{2} + 2 πn : θ = \frac{π}{4} + πn

Solve

2 θ = \frac{3 π}{2} + 2 πn : θ = \frac{3 π}{4} + πn

θ = \frac{π}{4} + πn, θ = \frac{3 π}{4} + πn

cos (x - \frac{π}{3}) = - 1

5 tan (B) - 11 = 0

2 csc (x) = 2

2 sin^{2} (x) + 2 cos (x) - 1 = 0

cos (x + π) = sin (x)

What is the general solution for 2sin^2(θ)-cos(2θ)=1 ?
The general solution for 2sin^2(θ)-cos(2θ)=1 is θ= pi/4+pin,θ=(3pi)/4+pin