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

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

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2sin(θ)tan(θ)-tan(θ)=0

2 sin (θ) tan (θ) - tan (θ) = 0

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

Solution steps

2 sin (θ) tan (θ) - tan (θ) = 0

Factor

2 sin (θ) tan (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

sec (x) + 2 = 0

sin (x) + 1 = 0, 0 \leq x \leq 2 π

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

12 tan (x) + 5 = 5 tan (x) + 5

4 cos (x) + 3 sin (x) = 0

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