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

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

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sin(θ)-cos(2θ)=0

sin (θ) - cos (2 θ) = 0

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

Solution steps

sin (θ) - cos (2 θ) = 0

Rewrite using trig identities

- 1 + sin (θ) + 2 sin^{2} (θ) = 0

Solve by substitution

sin (θ) = \frac{1}{2}, sin (θ) = - 1

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

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

Combine all the solutions

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

sec (θ) = - \frac{5}{3}

3 sin (x) = 2 cos (x)

5 cot (x) + 2 csc^{2} (x) = 5

cos (2 x) = \frac{4}{5}

tan (x) = \frac{160}{85}

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