# Solving equations of the form $\operatorname{trig function}(x) = \mathrm{constant}$

Is there a way to solve for x where $$\operatorname{trig function}(x) = \mathrm{constant}$$ and where the domain is such that the function has an inverse.

For example,

\begin{align*}\sin x &= \sqrt3 / 2,\\ \cos x &= -1,\\ \tan x &= 1/\sqrt3\end{align*}

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In general, you can't further simplify things like $\arcsin\left(\frac{\sqrt 2}{3}\right)$. However, for the second and third bits, it helps if you remember the sines and cosines of the more familiar angles... – J. M. Sep 23 '11 at 2:18
@J.M. Ok. $\sqrt2/3$ was actually supposed to be $\sqrt3/2$, so I'm guessing that is also a common angle. Is there some kind of connection between trigonometry and fractions with roots? Do you know where I might find a list of the common angles? – Matt Munson Sep 23 '11 at 2:25
Ah, then $\frac{\sqrt{3}}{2}$ is indeed a common angle. Formulae 3-7 here might help you with remembering those common angles... – J. M. Sep 23 '11 at 2:28
@Matt, De Moivre's formula is the source of a connection between trigonometry and roots. It's not completely direct though, and the question of which angles it works for is surprisingly nontrivial (and is one of the prime motivations for Galois theory). – Henning Makholm Sep 23 '11 at 2:36

## 1 Answer

All of the examples that you give (now that we have changed the first constant to $\sqrt{3}/2$ ) are values on the unit circle.
When you first learn about this, you'll be asked which angles satisfy the equation:

TrigFunction(angle) = constant, where the trig functions are:

sin, cos, tan, csc, sec, cot, and the constants are quotients involving:

$\pm \{0, \frac{1}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, 1 \}$

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(This might not have been too long for a comment, but I needed the compiler to help with my LaTexing) – The Chaz 2.0 Sep 23 '11 at 2:39