What is the general solution for cos(2x)+9cos(x)+5=0 ?

The general solution for cos(2x)+9cos(x)+5=0 is x=(2pi)/3+2pin,x=(4pi)/3+2pin

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cos(2x)+9cos(x)+5=0

cos (2 x) + 9 cos (x) + 5 = 0

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

Solution steps

cos (2 x) + 9 cos (x) + 5 = 0

Rewrite using trig identities

4 + 2 cos^{2} (x) + 9 cos (x) = 0

Solve by substitution

cos (x) = - \frac{1}{2}, cos (x) = - 4

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

cos (x) = - 4 :

No Solution

Combine all the solutions

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

25 = - 20 cos (\frac{2 π}{365} (x + 10)) + 25

sin (x) - cos (x) - \frac{1}{2} = 0

cos^{2} (θ) = \frac{49}{64}

sec^{2} (θ) = 1 - tan (θ)

cos^{2} (x) = - 0.267

What is the general solution for cos(2x)+9cos(x)+5=0 ?
The general solution for cos(2x)+9cos(x)+5=0 is x=(2pi)/3+2pin,x=(4pi)/3+2pin