What is the general solution for cos(2θ)-sin(θ)=0 ?

The general solution for cos(2θ)-sin(θ)=0 is θ=(3pi)/2+2pin,θ= pi/6+2pin,θ=(5pi)/6+2pin

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cos(2θ)-sin(θ)=0

cos (2 θ) - sin (θ) = 0

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

Solution steps

cos (2 θ) - sin (θ) = 0

Rewrite using trig identities

1 - sin (θ) - 2 sin^{2} (θ) = 0

Solve by substitution

sin (θ) = - 1, sin (θ) = \frac{1}{2}

sin (θ) = - 1 : θ = \frac{3 π}{2} + 2 πn

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

Combine all the solutions

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

tan (θ) = - \frac{2 3}{3} sin (θ)

3 cos (x) tan (x) - cos (x) = 0

4 sin^{2} (θ) = 3

cos (2 x) = sin (x) + 1

sec^{2} (x) - sec (x) - 2 = 0

What is the general solution for cos(2θ)-sin(θ)=0 ?
The general solution for cos(2θ)-sin(θ)=0 is θ=(3pi)/2+2pin,θ= pi/6+2pin,θ=(5pi)/6+2pin