What is the general solution for 2cos^2(θ)+5cos(θ)-2=-4 ?

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

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2cos^2(θ)+5cos(θ)-2=-4

2 cos^{2} (θ) + 5 cos (θ) - 2 = - 4

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

Solution steps

2 cos^{2} (θ) + 5 cos (θ) - 2 = - 4

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

tan (x) = \frac{2}{4}

7 cos (x) + 12 = 6 cos (x) + 13

sin (\frac{x}{2}) + cos (\frac{x}{2}) = 2

cos (θ) = - \frac{1}{2}, 0 \leq θ < 2 π

2 tan (x) - 16 = 0, 0 \leq x < \frac{π}{2}

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