# How do you solve #1+ cos x + cos 2x = 0#?

Use the identity for the cosine double angle formula:

The equation then becomes

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To solve the equation (1 + \cos(x) + \cos(2x) = 0), follow these steps:

- Use the double-angle identity for cosine: (\cos(2x) = 2\cos^2(x) - 1).
- Substitute the double-angle identity into the equation: (1 + \cos(x) + 2\cos^2(x) - 1 = 0).
- Simplify the equation: (2\cos^2(x) + \cos(x) = 0).
- Factor out common terms: (\cos(x)(2\cos(x) + 1) = 0).
- Set each factor equal to zero and solve for (x):
- (\cos(x) = 0) gives solutions where (x) is an odd multiple of (\frac{\pi}{2}).
- (2\cos(x) + 1 = 0) gives (\cos(x) = -\frac{1}{2}), which has solutions (x = \frac{2\pi}{3} + 2n\pi) and (x = \frac{4\pi}{3} + 2n\pi), where (n) is an integer.

These are the solutions to the equation (1 + \cos(x) + \cos(2x) = 0).

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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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