How would I solve #cos x + cos 2x = 0#? Please show steps.

Answer 1

We know that

#cos2x=cos^2x-sin^2x=cos^2x-(1-cos^2x)=2cos^2x-1#

hence the equation is

#cosx+cos2x=cosx+2cos^2x-1#

Hence we have to solve the

#2cos^2x+cosx-1=0=>(cosx+1)*(2cosx-1)=0#

or

#cosx=-1=>cosx=cospi=>x=2*k*pi+-pi#

and

#2cosx-1=0=>cosx=1/2=>cosx=cos(pi/3)=>x=2*k*pi+-pi/3#
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Answer 2

To solve the equation (\cos(x) + \cos(2x) = 0), we'll use trigonometric identities and algebraic manipulation.

First, recall the double angle identity for cosine: (\cos(2x) = 2\cos^2(x) - 1). Substitute this into the equation:

[ \cos(x) + 2\cos^2(x) - 1 = 0 ]

Rearrange terms:

[ 2\cos^2(x) + \cos(x) - 1 = 0 ]

Now, let (u = \cos(x)):

[ 2u^2 + u - 1 = 0 ]

This is now a quadratic equation. We can solve it using the quadratic formula:

[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ]

where (a = 2), (b = 1), and (c = -1).

Substitute these values into the quadratic formula and solve for (u). Then, substitute back (\cos(x)) for (u) to find the solutions for (x).

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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.

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.

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.

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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