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

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

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tan(2θ)-2sin(θ)=0

tan (2 θ) - 2 sin (θ) = 0

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

Solution steps

tan (2 θ) - 2 sin (θ) = 0

Express with sin, cos

\frac{sin ( 2 θ ) - 2 cos ( 2 θ ) sin ( θ )}{cos ( 2 θ )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

sin (2 θ) - 2 cos (2 θ) sin (θ) = 0

Rewrite using trig identities

- 2 cos (2 θ) sin (θ) + 2 cos (θ) sin (θ) = 0

Factor

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

2 sin (θ) (- cos (2 θ) + cos (θ)) = 0

Solving each part separately

sin (θ) = 0 or - cos (2 θ) + cos (θ) = 0

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

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

Combine all the solutions

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

sin (x) cos (x) = 3 cos (x)

sin (π x) = 0

csc (x - π) cos^{2} (x - π) - cot (x - π) = 0

tan^{2} (θ) + 6 tan (θ) + 8 = 0

sin (\frac{x}{2}) + cos (x) = 1

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