What is the general solution for sin^2(θ)=8(cos(-θ)+1) ?

The general solution for sin^2(θ)=8(cos(-θ)+1) is θ=pi+2pin

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sin^2(θ)=8(cos(-θ)+1)

sin^{2} (θ) = 8 (cos (- θ) + 1)

θ = π + 2 πn

Solution steps

sin^{2} (θ) = 8 (cos (- θ) + 1)

Rewrite using trig identities

sin^{2} (θ) = 8 (1 + cos (θ))

Subtract

8 (1 + cos (θ))

from both sides

sin^{2} (θ) - 8 (1 + cos (θ)) = 0

Rewrite using trig identities

- 7 - cos^{2} (θ) - 8 cos (θ) = 0

Solve by substitution

cos (θ) = - 7, cos (θ) = - 1

cos (θ) = - 7 :

No Solution

cos (θ) = - 1 : θ = π + 2 πn

Combine all the solutions

θ = π + 2 πn

tan (x) = \frac{7.1}{5}

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

sin (2 θ) = - \frac{1}{2}, θ \geq 0

tan (x) = \frac{1}{3}, 0 \leq x \leq 2 π

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

What is the general solution for sin^2(θ)=8(cos(-θ)+1) ?
The general solution for sin^2(θ)=8(cos(-θ)+1) is θ=pi+2pin