What is the general solution for 2-2sin(θ)=4cos^2(θ) ?

The general solution for 2-2sin(θ)=4cos^2(θ) is θ= pi/2+2pin,θ=(7pi)/6+2pin,θ=(11pi)/6+2pin

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2-2sin(θ)=4cos^2(θ)

2 - 2 sin (θ) = 4 cos^{2} (θ)

θ = \frac{π}{2} + 2 πn, θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Solution steps

2 - 2 sin (θ) = 4 cos^{2} (θ)

Subtract

4 cos^{2} (θ)

from both sides

2 - 2 sin (θ) - 4 cos^{2} (θ) = 0

Rewrite using trig identities

- 2 - 2 sin (θ) + 4 sin^{2} (θ) = 0

Solve by substitution

sin (θ) = 1, sin (θ) = - \frac{1}{2}

sin (θ) = 1 : θ = \frac{π}{2} + 2 πn

sin (θ) = - \frac{1}{2} : θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Combine all the solutions

θ = \frac{π}{2} + 2 πn, θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

cos (θ) = 0.5678

1 = cos (2 x)

tan (x) = 2 cos (x) tan (x)

tan (θ) = 2.6

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

What is the general solution for 2-2sin(θ)=4cos^2(θ) ?
The general solution for 2-2sin(θ)=4cos^2(θ) is θ= pi/2+2pin,θ=(7pi)/6+2pin,θ=(11pi)/6+2pin