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

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

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2sin(θ)cos(θ)+2sin(θ)-2cos(θ)sin(θ)=0

2 sin (θ) cos (θ) + 2 sin (θ) - 2 cos (θ) sin (θ) = 0

θ = 2 πn, θ = π + 2 πn

Solution steps

2 sin (θ) cos (θ) + 2 sin (θ) - 2 cos (θ) sin (θ) = 0

Simplify

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

2 sin (θ) = 0

Divide both sides by

2

sin (θ) = 0

General solutions for

sin (θ) = 0

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 2 πn, θ = π + 2 πn

2 sin (2 x) + 2 cos (x) = 0

cosh (x) = \frac{3}{2}

\frac{1 + tan ( t )}{sin ( t )} = 0

sin^{2} (θ) + 2 cos (θ) = \frac{7}{4}

5 cos (x) + 3 = 3 cos (x) + 5

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