How can trigonometric identities be used to simplify complex expressions?

[zeion]

Homework Statement

I need to solve this:

cosx - cos2x + cos3x = 0

Homework Equations

The Attempt at a Solution

The solution shows something like:
(cosx + cos3x) - cos2x

Then using the sum to product on the first group and double angle on the cos2x.

I would know to use double angle identity somewhere, but I would not think of grouping the first 2 things. Just wondering if there's an easier way? Or what kind of mindset or goal should I have to solve something like this? It seems rather arbitrary.
Thanks.

 

[Metaleer]

Remembering that $$\cos 2x = 2 \cos^2 x - 1$$ and that $$\cos 3x = 4 \cos^3 x - 3 \cos x$$ (the proofs for these identities are not difficult), your equation becomes

$$\cos x - (2 \cos^2 x - 1) + 4 \cos^3 x - 3 \cos x = 0$$​

and you do a substitution, $$\cos x = t$$ which yields a polynomial equation in $$t$$, you solve this equation and then you take the inverse of $$\cos x$$, $$\arccos t$$ to get your angles.

 

[zeion]

Remembering that $$\cos 2x = 2 \cos^2 x - 1$$ and that $$\cos 3x = 4 \cos^3 x - 3 \cos x$$ (the proofs for these identities are not difficult), your equation becomes

$$\cos x - (2 \cos^2 x - 1) + 4 \cos^3 x - 3 \cos x = 0$$​

and you do a substitution, $$\cos x = t$$ which yields a polynomial equation in $$t$$, you solve this equation and then you take the inverse of $$\cos x$$, $$\arccos t$$ to get your angles.

I don't think that's much easier than the solution I posted...
I haven't covered inverse of trig functions yet at this point of the book. Only the major trig identities.

 

[Metaleer]

Have a go at it. I just solved it and the results that show up are angles that you need to know by heart.

 

[eumyang]

I don't think that's much easier than the solution I posted...
I haven't covered inverse of trig functions yet at this point of the book. Only the major trig identities.

But if you are given some of the trig ratios you should be able to get the angles. For instance, if I were to ask you to find theta if
$$\sin \theta = \frac{1}{2}$$
and 0 ≤ θ < 2π, then you should be able to tell me that
$$\theta = \frac{\pi}{6}, \frac{5\pi}{6}$$.

BTW, the cubic that Metaleer gave factors pretty easily.