What is the general solution for 2cos^2(θ)+5cos(θ)+2=0 ?

The general solution for 2cos^2(θ)+5cos(θ)+2=0 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin

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2cos^2(θ)+5cos(θ)+2=0

2 cos^{2} (θ) + 5 cos (θ) + 2 = 0

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

Solution steps

2 cos^{2} (θ) + 5 cos (θ) + 2 = 0

Solve by substitution

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

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

cos (θ) = - 2 :

No Solution

Combine all the solutions

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

sec^{2} (x) + tan (x) = 0

sin (θ) cot^{2} (θ) - 3 sin (θ) = 0

- 6 sin (x) - 3 = 9

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

cos (θ) = - \frac{3}{7}

What is the general solution for 2cos^2(θ)+5cos(θ)+2=0 ?
The general solution for 2cos^2(θ)+5cos(θ)+2=0 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin