How do you solve cos^2x+6cosx+4=0cos2x+6cosx+4=0 in the interval [0,360]?

1 Answer
Alberto P.
Oct 26, 2016

arccos(-3+sqrt(5))arccos(−3+√5)

Explanation:

Let Y=cos xY=cosx then
Y^2+6Y+4=0 and abs(Y)<=1Y2+6Y+4=0and|Y|≤1
Solving the equation
Y_(1,2)=-3+-sqrt(9-4)=-3+-sqrt(5)Y1,2=−3±√9−4=−3±√5
Only the solution with + satisfy abs(Y)>=1|Y|≥1

So

cos x = -3+sqrt(5)cosx=−3+√5

and

arccos(-3+sqrt(5))arccos(−3+√5)

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