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

The general solution for sin(2θ)=cos(θ) is θ= pi/2+2pin,θ=(3pi)/2+2pin,θ= pi/6+2pin,θ=(5pi)/6+2pin

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sin(2θ)=cos(θ)

sin (2 θ) = cos (θ)

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn, θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Solution steps

sin (2 θ) = cos (θ)

Subtract

cos (θ)

from both sides

sin (2 θ) - cos (θ) = 0

Rewrite using trig identities

- cos (θ) + 2 cos (θ) sin (θ) = 0

Factor

- cos (θ) + 2 cos (θ) sin (θ) : cos (θ) (2 sin (θ) - 1)

cos (θ) (2 sin (θ) - 1) = 0

Solving each part separately

cos (θ) = 0 or 2 sin (θ) - 1 = 0

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

2 sin (θ) - 1 = 0 : θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Combine all the solutions

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn, θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

sin (θ) = \frac{2}{3}

sin (2 θ) = 0

tan (x) = \frac{3}{2}

cos (4 x) = 0

sin^{3} (x) = sin (x)

What is the general solution for sin(2θ)=cos(θ) ?
The general solution for sin(2θ)=cos(θ) is θ= pi/2+2pin,θ=(3pi)/2+2pin,θ= pi/6+2pin,θ=(5pi)/6+2pin