# Simple trig problem

1. Oct 4, 2015

### Mddrill

1. The problem statement, all variables and given/known data
I feel like I this problem shouldn't be that hard, but I cant figure out how to evaluate this equation

cosx - cosx*sinx = 1/3

2. Relevant equations
I don't know what to use for this, none of the trig identities seem to help

3. The attempt at a solution
I tried substituting $sqrt(1-cos^2x )$ for sinx , but I just ended up with something way more complicated. I know this shouldn't be this hard because I'm not even in a trig class, Im taking statics. It wouldn't make sense for them to give me a problem that requires more time with the trig than with the statics.

Thank You

2. Oct 4, 2015

### Staff: Mentor

WolframAlpha.com shows solutions to this, but it doesn't seem to show a simplified trig form...

3. Oct 4, 2015

### Ray Vickson

This is simple to solve numerically, but not at all easy to solve "analytically" (i.e., using formulas). For what it's worth, Maple 11 obtains the following "analytic" solution:
x =
arctan((-1+3*(1/6*(((162+6*681^(1/2))^(2/3)+12)/(162+6*681^(1/2))^(1/3))^(1/2)+1/6*((-(((162+6*681^(1/2))^(2/3)+12)/(162+6*681^(1/2))^(1/3))^(1/2)*(162+6*681^(1/2))^(2/3)-12*(((162+6*681^(1/2))^(2/3)+12)/(162+6*681^(1/2))^(1/3))^(1/2)+36*(162+6*681^(1/2))^(1/3))/(162+6*681^(1/2))^(1/3)/(((162+6*681^(1/2))^(2/3)+12)/(162+6*681^(1/2))^(1/3))^(1/2))^(1/2))^3)/(1/6*(((162+6*681^(1/2))^(2/3)+12)/(162+6*681^(1/2))^(1/3))^(1/2)+1/6*((-(((162+6*681^(1/2))^(2/3)+12)/(162+6*681^(1/2))^(1/3))^(1/2)*(162+6*681^(1/2))^(2/3)-12*(((162+6*681^(1/2))^(2/3)+12)/(162+6*681^(1/2))^(1/3))^(1/2)+36*(162+6*681^(1/2))^(1/3))/(162+6*681^(1/2))^(1/3)/(((162+6*681^(1/2))^(2/3)+12)/(162+6*681^(1/2))^(1/3))^(1/2))^(1/2)))
Basically, this has the form
$$x = \arctan\left( \frac{-1 + 3z^3}{z} \right),$$
where $z$ is a root of the equation $9z^4 -6z + 1 = 0$. Numerically, $x \doteq 0.6284918352$. Of course, there are other solutions outside the region $0 < x < \pi$.