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

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

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1-2sin(θ)=cos(2θ)

1 - 2 sin (θ) = cos (2 θ)

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

Solution steps

1 - 2 sin (θ) = cos (2 θ)

Subtract

cos (2 θ)

from both sides

1 - 2 sin (θ) - cos (2 θ) = 0

Rewrite using trig identities

- 2 sin (θ) + 2 sin^{2} (θ) = 0

Solve by substitution

sin (θ) = 1, sin (θ) = 0

sin (θ) = 1 : θ = \frac{π}{2} + 2 πn

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

Combine all the solutions

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

cos (π)

sec^{2} (1)

x, cos^{2} (x) - sin^{2} (x) = sin (x)

arcsin (2)

cos (4 x) = sin (x)

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