In Problem 1 of the Solving Trig Equations section we solved the following equation.

\[\cos \left( t \right) = \frac{{\sqrt 3 }}{2}\]

In other words, we asked what angles, \(x\), do we need to plug into cosine to get \(\frac{{\sqrt 3 }}{2}\)? This is essentially what we are asking here when we are asked to compute the inverse trig function.

\[{\cos ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right)\]

There is one very large difference however. In Problem 1 we were solving an equation which yielded an infinite number of solutions. These were,

\[\begin{align*} & \frac{\pi }{6} + 2\pi \,n\,,\;\;n = 0,\, \pm 1,\, \pm 2,\, \pm 3,\, \ldots \\ & \frac{{11\pi }}{6} + 2\pi \,n\,,\;\;n = 0,\, \pm 1,\, \pm 2,\, \pm 3,\, \ldots \end{align*}\]

In the case of inverse trig functions, we are after a single value. We don’t want to have to guess at which one of the infinite possible answers we want. So, to make sure we get a single value out of the inverse trig cosine function we use the following restrictions on inverse cosine.

\[\theta = {\cos ^{ - 1}}\left( x \right)\hspace{0.25in}\hspace{0.25in} - 1 \le x \le 1\hspace{0.25in}{\rm{and }}\hspace{0.25in}0 \le \theta \le \pi \]

The restriction on the \(\theta \) guarantees that we will only get a single value angle and since we can’t get values of \(x\) out of cosine that are larger than 1 or smaller than -1 we also can’t plug these values into an inverse trig function.

So, using these restrictions on the solution to Problem 1 we can see that the answer in this case is

\[{\cos ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right) = \frac{\pi }{6}\]