Question

How do you solve `cos^3x-2sinx-0.7=0`?

Answer 1

We cannot exactly solve it, but there are approximate solutions at `-2.5 + 2pi n and 0.14 + 2pi n` for `n in mathbb{Z}`.

Explanation:

We cannot exactly solve the equation with the given values.

What we want to do, generally, is combine trigonometric terms using identities. Here, there aren't any useful identities, even if we start breaking things up.

If we plot the function,
graph{-0.7-2sin(x) + cos(x)^3 [-10, 10, -5, 5]}

we see that it is periodic with period `2pi`, but none of those roots are analytically expressible except as "the solution to this function".

I'll go into a little nitty gritty here, using complex exponentials:
We can let
`cos(x) = 1/2 (e^(ix) + e^(-ix))` and `sin(x) = 1/(2i)(e^(ix)-e^(-ix))`\ With `u = e^(ix)`, this means our equation becomes
`1/8(u + 1/u)^3 - 1/i (u - 1/u) = 0.7`
`(u^2 + 1)^3 + 8i (u^4 - u^2) = 5.6u^3`
`u^6 + 3u^4 + 3u^2 + 1 + 8iu^4 - 8iu^2 = 5.6 u^3`
`u^6 + (3+8i)u^4 - 5.6 u^3 + (3-8i)u^2 +1 = 0`

So we have a 6th order complex polynomial in terms of `e^(ix)`. We can solve it and it turns out that there's a set of solutions which have magnitude 1 which gives purely real values of `x`. We cannot analytically solve a sixth order polynomial, so we can only find numerical answers, such as with Newton's method.