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

Answer 1

#arccos(-3+sqrt(5))#

Let #Y=cos x# then #Y^2+6Y+4=0 and abs(Y)<=1# Solving the equation #Y_(1,2)=-3+-sqrt(9-4)=-3+-sqrt(5)# Only the solution with + satisfy #abs(Y)>=1#

#cos x = -3+sqrt(5)#

and

#arccos(-3+sqrt(5))#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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