What is the general solution for 2cos^2(θ)+3cos(θ)+1=0 ?

The general solution for 2cos^2(θ)+3cos(θ)+1=0 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin,θ=pi+2pin

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2cos^2(θ)+3cos(θ)+1=0

2 cos^{2} (θ) + 3 cos (θ) + 1 = 0

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn, θ = π + 2 πn

Solution steps

2 cos^{2} (θ) + 3 cos (θ) + 1 = 0

Solve by substitution

cos (θ) = - \frac{1}{2}, cos (θ) = - 1

cos (θ) = - \frac{1}{2} : θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

cos (θ) = - 1 : θ = π + 2 πn

Combine all the solutions

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn, θ = π + 2 πn

tan^{2} (x) sin (x) = - sin (x)

cos (4 x) = sin (x), 0 \leq x \leq 2 π

2 cos (x) = 1

sin (θ) = \frac{- 3}{2}

2 cos (\frac{x}{5}) - 2 = 0

What is the general solution for 2cos^2(θ)+3cos(θ)+1=0 ?
The general solution for 2cos^2(θ)+3cos(θ)+1=0 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin,θ=pi+2pin