What is the general solution for cos^2(x)+4cos(x)+1=0 ?

The general solution for cos^2(x)+4cos(x)+1=0 is x=1.84206…+2pin,x=-1.84206…+2pin

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cos^2(x)+4cos(x)+1=0

cos^{2} (x) + 4 cos (x) + 1 = 0

x = 1.84206 \dots + 2 πn, x = - 1.84206 \dots + 2 πn

Solution steps

cos^{2} (x) + 4 cos (x) + 1 = 0

Solve by substitution

cos (x) = - 2 + 3, cos (x) = - 2 - 3

cos (x) = - 2 + 3 : x = arccos (- 2 + 3) + 2 πn, x = - arccos (- 2 + 3) + 2 πn

cos (x) = - 2 - 3 :

No Solution

Combine all the solutions

x = arccos (- 2 + 3) + 2 πn, x = - arccos (- 2 + 3) + 2 πn

Show solutions in decimal form

x = 1.84206 \dots + 2 πn, x = - 1.84206 \dots + 2 πn

8 cos (2 θ) = - 8 cos (θ)

arcsin (x) = \frac{π}{4}

arcsin (x) = \frac{π}{3}

sin^{2} (x) = 4

6 - sin (θ) = cos (2 θ)

What is the general solution for cos^2(x)+4cos(x)+1=0 ?
The general solution for cos^2(x)+4cos(x)+1=0 is x=1.84206…+2pin,x=-1.84206…+2pin