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

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

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sin(θ)-2cos^2(θ)=-1

sin (θ) - 2 cos^{2} (θ) = - 1

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

Solution steps

sin (θ) - 2 cos^{2} (θ) = - 1

Subtract

- 1

from both sides

sin (θ) - 2 cos^{2} (θ) + 1 = 0

Rewrite using trig identities

- 1 + sin (θ) + 2 sin^{2} (θ) = 0

Solve by substitution

sin (θ) = \frac{1}{2}, sin (θ) = - 1

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

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

Combine all the solutions

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

\frac{1}{1 - cos ( x )} + \frac{1}{1 + cos ( x )} = 2 csc (x)

\frac{3}{8} = cos (θ)

cos (θ) = 0.75

cos (χ) = 32

x, tan^{3} (x) = f

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