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

The general solution for 2cos^2(θ)-sin(θ)=1 is θ=(3pi)/2+2pin,θ= pi/6+2pin,θ=(5pi)/6+2pin

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2cos^2(θ)-sin(θ)=1

2 cos^{2} (θ) - sin (θ) = 1

θ = \frac{3 π}{2} + 2 πn, θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Solution steps

2 cos^{2} (θ) - sin (θ) = 1

Subtract

1

from both sides

2 cos^{2} (θ) - sin (θ) - 1 = 0

Rewrite using trig identities

1 - sin (θ) - 2 sin^{2} (θ) = 0

Solve by substitution

sin (θ) = - 1, sin (θ) = \frac{1}{2}

sin (θ) = - 1 : θ = \frac{3 π}{2} + 2 πn

sin (θ) = \frac{1}{2} : θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Combine all the solutions

θ = \frac{3 π}{2} + 2 πn, θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

cosh (x) = 0

cos (x) = \frac{5}{13}

3 sin (x) + cos (x) = 1

sin (t) = \frac{1}{2}

sin (θ) - cos (θ) = 2

What is the general solution for 2cos^2(θ)-sin(θ)=1 ?
The general solution for 2cos^2(θ)-sin(θ)=1 is θ=(3pi)/2+2pin,θ= pi/6+2pin,θ=(5pi)/6+2pin