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

The general solution for sin(θ)cos(2θ)-cos(2θ)=0 is θ= pi/4+pin,θ=(3pi)/4+pin,θ= pi/2+2pin

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sin(θ)cos(2θ)-cos(2θ)=0

sin (θ) cos (2 θ) - cos (2 θ) = 0

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

Solution steps

sin (θ) cos (2 θ) - cos (2 θ) = 0

Factor

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

cos (2 θ) (sin (θ) - 1) = 0

Solving each part separately

cos (2 θ) = 0 or sin (θ) - 1 = 0

cos (2 θ) = 0 : θ = \frac{π}{4} + πn, θ = \frac{3 π}{4} + πn

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

Combine all the solutions

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

tan (θ) = - \frac{7}{6}

8 sin^{2} (x) - 1 = 5

sin (d) = 0.8

csc (θ) = \frac{17}{8}

\frac{π}{3} cos (\frac{π x}{6}) = 0, x

What is the general solution for sin(θ)cos(2θ)-cos(2θ)=0 ?
The general solution for sin(θ)cos(2θ)-cos(2θ)=0 is θ= pi/4+pin,θ=(3pi)/4+pin,θ= pi/2+2pin