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

The general solution for 2sin^2(θ)=cos^2(θ) is θ=-0.61547…+pin,θ=0.61547…+pin

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2sin^2(θ)=cos^2(θ)

2 sin^{2} (θ) = cos^{2} (θ)

θ = - 0.61547 \dots + πn, θ = 0.61547 \dots + πn

Solution steps

2 sin^{2} (θ) = cos^{2} (θ)

Subtract

cos^{2} (θ)

from both sides

2 sin^{2} (θ) - cos^{2} (θ) = 0

Factor

2 sin^{2} (θ) - cos^{2} (θ) : (2 sin (θ) + cos (θ)) (2 sin (θ) - cos (θ))

(2 sin (θ) + cos (θ)) (2 sin (θ) - cos (θ)) = 0

Solving each part separately

2 sin (θ) + cos (θ) = 0 or 2 sin (θ) - cos (θ) = 0

2 sin (θ) + cos (θ) = 0 : θ = arctan (- \frac{2}{2}) + πn

2 sin (θ) - cos (θ) = 0 : θ = arctan (\frac{2}{2}) + πn

Combine all the solutions

θ = arctan (- \frac{2}{2}) + πn, θ = arctan (\frac{2}{2}) + πn

Show solutions in decimal form

θ = - 0.61547 \dots + πn, θ = 0.61547 \dots + πn

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

x, y + arctan (\frac{y}{x}) = 0

tan (θ) = - 1, 0 \leq θ \leq 2 π

0.85 = \frac{1 - sin ( x )}{1 + sin ( x )}

sin (2 x) - \frac{2}{2} = 0

What is the general solution for 2sin^2(θ)=cos^2(θ) ?
The general solution for 2sin^2(θ)=cos^2(θ) is θ=-0.61547…+pin,θ=0.61547…+pin