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

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

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sqrt(2)sin(θ)=sqrt(cos(2θ))

2 sin (θ) = cos (2 θ)

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

Solution steps

2 sin (θ) = cos (2 θ)

Subtract

cos (2 θ)

from both sides

2 sin (θ) - cos (2 θ) = 0

Rewrite using trig identities

- 1 - 2 sin^{2} (θ) + sin (θ) 2 = 0

Solve by substitution

sin (θ) = \frac{1}{2}

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

Combine all the solutions

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

tan (θ) = \frac{4}{- 4}

sin (x) = \frac{2}{7}

sin (θ) cos (θ) + sin (θ) = 0

6 cos^{2} (x) - cos (x) - 1 = 0

45.6 = arcsin (\frac{n ^{1}}{n ^{2}}), n^{1}

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