What is the general solution for 2cos^2(2θ)=1-cos(2θ) ?

The general solution for 2cos^2(2θ)=1-cos(2θ) is θ= pi/6+pin,θ=(5pi)/6+pin,θ= pi/2+pin

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2cos^2(2θ)=1-cos(2θ)

2 cos^{2} (2 θ) = 1 - cos (2 θ)

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

Solution steps

2 cos^{2} (2 θ) = 1 - cos (2 θ)

Solve by substitution

cos (2 θ) = \frac{1}{2}, cos (2 θ) = - 1

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

cos (2 θ) = - 1 : θ = \frac{π}{2} + πn

Combine all the solutions

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

1 + tan^{2} (x) = 0

sin (x) = \frac{3}{8}

x, tan (x) + cot (x) = \frac{1}{f}

sec^{2} (x) - 1 + tan (x) - 3 tan (x) = 3

3 sec (2 x) - 2 = 0

What is the general solution for 2cos^2(2θ)=1-cos(2θ) ?
The general solution for 2cos^2(2θ)=1-cos(2θ) is θ= pi/6+pin,θ=(5pi)/6+pin,θ= pi/2+pin