What is the general solution for -2cos^2(θ)+sin(θ)-3=-4 ?

The general solution for -2cos^2(θ)+sin(θ)-3=-4 is θ= pi/6+2pin,θ=(5pi)/6+2pin,θ=(3pi)/2+2pin

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-2cos^2(θ)+sin(θ)-3=-4

- 2 cos^{2} (θ) + sin (θ) - 3 = - 4

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

Solution steps

- 2 cos^{2} (θ) + sin (θ) - 3 = - 4

Subtract

- 4

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

sec^{2} (x) - 9 sec (x) = 0

- 3 sin (2 θ) = 5 cos (2 θ)

(tan^{2} (θ) - 4) (2 cos (θ) + 1) = 0

cot (t) = - 1

cot^{2} (4 x) = 3

What is the general solution for -2cos^2(θ)+sin(θ)-3=-4 ?
The general solution for -2cos^2(θ)+sin(θ)-3=-4 is θ= pi/6+2pin,θ=(5pi)/6+2pin,θ=(3pi)/2+2pin