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

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

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2sin(θ)cos(θ)=sin(θ)

2 sin (θ) cos (θ) = sin (θ)

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

Solution steps

2 sin (θ) cos (θ) = sin (θ)

Subtract

sin (θ)

from both sides

2 sin (θ) cos (θ) - sin (θ) = 0

Factor

2 sin (θ) cos (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

53 cot (x) + 5 = 0

cos (θ) = - \frac{11}{6}

sin^{2} (x) - csc^{2} (x) = tan^{2} (x) - cot^{2} (x)

sin (x) = \frac{53}{54}

2 sin^{2} (x) + sin (x) - 3 = 0

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