What is the general solution for 2cos(θ)sin(θ)+cos(θ)-2sin(θ)=1 ?

The general solution for 2cos(θ)sin(θ)+cos(θ)-2sin(θ)=1 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin,θ=2pin

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2cos(θ)sin(θ)+cos(θ)-2sin(θ)=1

2 cos (θ) sin (θ) + cos (θ) - 2 sin (θ) = 1

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn, θ = 2 πn

Solution steps

2 cos (θ) sin (θ) + cos (θ) - 2 sin (θ) = 1

Subtract

1

from both sides

2 cos (θ) sin (θ) + cos (θ) - 2 sin (θ) - 1 = 0

Factor

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

(2 sin (θ) + 1) (cos (θ) - 1) = 0

Solving each part separately

2 sin (θ) + 1 = 0 or cos (θ) - 1 = 0

2 sin (θ) + 1 = 0 : θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

cos (θ) - 1 = 0 : θ = 2 πn

Combine all the solutions

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn, θ = 2 πn

sin (2 θ) - cos (θ) - 2 sin (θ) + 1 = 0

2 cos (x) + 1 = 0 - 2, π \leq x \leq 2 π

cos^{3} (x) = - cos (x)

csc (2 x) = 2 sin (2 x), 0 \leq x \leq 2 π

cos (x) = - \frac{12}{13}

What is the general solution for 2cos(θ)sin(θ)+cos(θ)-2sin(θ)=1 ?
The general solution for 2cos(θ)sin(θ)+cos(θ)-2sin(θ)=1 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin,θ=2pin