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

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

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sin^2(θ)=2cos(θ)+1

sin^{2} (θ) = 2 cos (θ) + 1

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

Solution steps

sin^{2} (θ) = 2 cos (θ) + 1

Subtract

2 cos (θ) + 1

from both sides

sin^{2} (θ) - 2 cos (θ) - 1 = 0

Rewrite using trig identities

- cos^{2} (θ) - 2 cos (θ) = 0

Solve by substitution

cos (θ) = - 2, cos (θ) = 0

cos (θ) = - 2 :

No Solution

cos (θ) = 0 : θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Combine all the solutions

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

sin (6 x) cos (10 x) - cos (6 x) sin (10 x) = - 0.1

(cos (θ)) (cos (θ) + 1) = 0

5 - tan^{2} (x) = 4, 0^{\circ} \leq x \leq 36 0^{\circ}

- 2 tan (x) + 1 = 23 + 1

sin (x - \frac{π}{3}) = \frac{1}{2}

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