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

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

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2cos(θ)sin(θ)-sin(θ)=0

2 cos (θ) sin (θ) - sin (θ) = 0

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

Solution steps

2 cos (θ) sin (θ) - sin (θ) = 0

Factor

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

sin (θ) (2 cos (θ) - 1) = 0

Solving each part separately

sin (θ) = 0 or 2 cos (θ) - 1 = 0

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

2 cos (θ) - 1 = 0 : θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

Combine all the solutions

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

2 sin (x) cot (x) - cot (x) = - 1 + 2 sin (x)

sin (3 θ) = cos (3 θ)

2 + sec (x) = 0

cos (x) = \frac{3}{2}, sin (x), \frac{3 π}{2} \leq x \leq 2 π

sec (x) = 9

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