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

The general solution for 2cos^2(θ)=2-sin(θ) is θ=2pin,θ=pi+2pin,θ= pi/6+2pin,θ=(5pi)/6+2pin

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2cos^2(θ)=2-sin(θ)

2 cos^{2} (θ) = 2 - sin (θ)

θ = 2 πn, θ = π + 2 πn, θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Solution steps

2 cos^{2} (θ) = 2 - sin (θ)

Subtract

2 - sin (θ)

from both sides

2 cos^{2} (θ) - 2 + sin (θ) = 0

Rewrite using trig identities

sin (θ) - 2 sin^{2} (θ) = 0

Solve by substitution

sin (θ) = 0, sin (θ) = \frac{1}{2}

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

sin (θ) = \frac{1}{2} : θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Combine all the solutions

θ = 2 πn, θ = π + 2 πn, θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

cos (x + 1) = 0

sec^{2} (x) + tan^{2} (x) = 7

4 cos (x) = 1, 0 \leq x \leq 2 π

2 cos (2 x) + 4 sin (x) + 1 = 0

6 sin^{2} (θ) - 17 sin (θ) + 14 = - 4 sin (θ) + 9

What is the general solution for 2cos^2(θ)=2-sin(θ) ?
The general solution for 2cos^2(θ)=2-sin(θ) is θ=2pin,θ=pi+2pin,θ= pi/6+2pin,θ=(5pi)/6+2pin