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

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

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2cos^2(x)+5cos(x)+2=0

2 cos^{2} (x) + 5 cos (x) + 2 = 0

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

Solution steps

2 cos^{2} (x) + 5 cos (x) + 2 = 0

Solve by substitution

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

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

cos (x) = - 2 :

No Solution

Combine all the solutions

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

sin (θ) = - 0.66

4 sec^{2} (θ) tan (θ) = 8 tan (θ)

5 sec (x) csc (x) = 10 csc (x)

cos (x) = 1 - sin (x)

2 sin (2 θ) = cos (2 θ), 0 \leq θ \leq 2 π

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