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

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

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2sin(θ)cos(θ)-sin(θ)=sin(θ)cos(θ)

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

θ = 2 πn, θ = π + 2 πn

Solution steps

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

Subtract

sin (θ) cos (θ)

from both sides

sin (θ) cos (θ) - sin (θ) = 0

Factor

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

sin (θ) (cos (θ) - 1) = 0

Solving each part separately

sin (θ) = 0 or cos (θ) - 1 = 0

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

cos (θ) - 1 = 0 : θ = 2 πn

Combine all the solutions

θ = 2 πn, θ = π + 2 πn

3 - \frac{1}{2} \cdot tan (θ) = \frac{6 + 3}{2}

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

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

20 sin (t) cos (t) = - 4 sin (t)

cos (x) - cos (2 x) = 0, 0^{\circ} \leq x < 36 0^{\circ}

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