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

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

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2cos(θ)sin(θ)=sin(θ)

2 cos (θ) sin (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 cos (θ) sin (θ) = sin (θ)

Subtract

sin (θ)

from both sides

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

4 cos (x) = sec (x)

9.6 = 11.6 cos (3.922 t)

2 sin^{2} (x) = tan^{2} (x)

cos (θ) = 0.5632

sin (z) = - 10

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