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

The general solution for sin(2θ)=2cos^2(θ) is θ= pi/2+2pin,θ=(3pi)/2+2pin,θ= pi/4+pin

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sin(2θ)=2cos^2(θ)

sin (2 θ) = 2 cos^{2} (θ)

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn, θ = \frac{π}{4} + πn

Solution steps

sin (2 θ) = 2 cos^{2} (θ)

Subtract

2 cos^{2} (θ)

from both sides

sin (2 θ) - 2 cos^{2} (θ) = 0

Rewrite using trig identities

- 2 cos^{2} (θ) + 2 cos (θ) sin (θ) = 0

Factor

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

2 cos (θ) (- cos (θ) + sin (θ)) = 0

Solving each part separately

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

cos (θ) = 0 : θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

- cos (θ) + sin (θ) = 0 : θ = \frac{π}{4} + πn

Combine all the solutions

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn, θ = \frac{π}{4} + πn

0 = 3 cos (3 θ)

sin (\frac{1}{2} x) = cos (x)

tan (2 x - \frac{π}{4}) = - 1

2 sin^{2} (A) = 5 sin (A) - 2

cot (θ) = \frac{12}{5}

What is the general solution for sin(2θ)=2cos^2(θ) ?
The general solution for sin(2θ)=2cos^2(θ) is θ= pi/2+2pin,θ=(3pi)/2+2pin,θ= pi/4+pin