What is the general solution for 1-2sin(θ)=cos(2θ),0<= θ<= 2pi ?

The general solution for 1-2sin(θ)=cos(2θ),0<= θ<= 2pi is θ= pi/2 ,θ=0,θ=pi,θ=2pi

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1-2sin(θ)=cos(2θ),0<= θ<= 2pi

1 - 2 sin (θ) = cos (2 θ), 0 \leq θ \leq 2 π

θ = \frac{π}{2}, θ = 0, θ = π, θ = 2 π

Solution steps

1 - 2 sin (θ) = cos (2 θ), 0 \leq θ \leq 2 π

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, 0 \leq θ \leq 2 π : θ = \frac{π}{2}

sin (θ) = 0, 0 \leq θ \leq 2 π : θ = 0, θ = π, θ = 2 π

Combine all the solutions

θ = \frac{π}{2}, θ = 0, θ = π, θ = 2 π

2 sin^{2} (θ) + 5 sin (θ) - 12 = 0

cos (2 θ) = \frac{3}{5}, 9 0^{\circ} \leq θ \leq 18 0^{\circ}

tan (θ) = \frac{1}{- 1}

sin^{2} (x) - 4 cos (x) - 4 = 0, 0 \leq x \leq 2 π

cos^{2} (θ) = 2 - 2 sin (θ)

What is the general solution for 1-2sin(θ)=cos(2θ),0<= θ<= 2pi ?
The general solution for 1-2sin(θ)=cos(2θ),0<= θ<= 2pi is θ= pi/2 ,θ=0,θ=pi,θ=2pi