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

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

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tan(θ)=2sin(θ)tan(θ)

tan (θ) = 2 sin (θ) tan (θ)

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

Solution steps

tan (θ) = 2 sin (θ) tan (θ)

Subtract

2 sin (θ) tan (θ)

from both sides

tan (θ) - 2 sin (θ) tan (θ) = 0

Factor

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

- tan (θ) (2 sin (θ) - 1) = 0

Solving each part separately

tan (θ) = 0 or 2 sin (θ) - 1 = 0

tan (θ) = 0 : θ = πn

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

Combine all the solutions

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

2 sin^{2} (t) + sin (t) - 1 = 0

sin (θ) + sin (3 θ) = 0

4 sin^{2} (x) - 1 = 0, 0 \leq x < 2 π

cos^{2} (x) - 4 sin^{2} (x) = 2

sin (x) = 0.53

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