What is the general solution for cos^2(θ)+2cos(θ)+1=0,0<= θ<2pi ?

The general solution for cos^2(θ)+2cos(θ)+1=0,0<= θ<2pi is θ=pi

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cos^2(θ)+2cos(θ)+1=0,0<= θ<2pi

cos^{2} (θ) + 2 cos (θ) + 1 = 0, 0 \leq θ < 2 π

θ = π

Solution steps

cos^{2} (θ) + 2 cos (θ) + 1 = 0, 0 \leq θ < 2 π

Solve by substitution

cos (θ) = - 1

cos (θ) = - 1, 0 \leq θ < 2 π : θ = π

Combine all the solutions

θ = π

950 sin (\frac{π}{6} (7 - x)) + 1650 = 2500

sin (θ) = 12

sin (2 x - \frac{π}{6}) = - \frac{1}{2}

cos (x) (2 sin (x) + 3) = 0

cot (θ) + 2 csc (θ) = 4

What is the general solution for cos^2(θ)+2cos(θ)+1=0,0<= θ<2pi ?
The general solution for cos^2(θ)+2cos(θ)+1=0,0<= θ<2pi is θ=pi