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

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

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2tan(θ)sin(θ)-tan(θ)=0

2 tan (θ) sin (θ) - tan (θ) = 0

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

Solution steps

2 tan (θ) sin (θ) - tan (θ) = 0

Factor

2 tan (θ) 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

sin (4 x + \frac{π}{4}) = 1

cos^{2} (x) - cos (x) = 0, (0, 2 π)

csc (4 2^{\circ}) = sec (x)

x, 0 = cos (x) + 1 + sin (x)

5 sin (\frac{π}{4} x) = 4

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