A querry regarding inverse trig. functions

Question

Find the value of $\cos^{-1}\left({{-\sqrt 3}\over 2}\right)$

Let $\cos^{-1}\left({{-\sqrt 3}\over 2}\right) = y$

$\Rightarrow$ $\cos y = {{-\sqrt3}\over 2}$

I) $\cos y=\cos\left({-\pi\over6}\right)$
$\cos y=\cos\left(\pi - {\pi\over6}\right)$
$\Rightarrow$ $\cos y=\cos\left({5\pi\over6}\right)$
$\Rightarrow$ $y={5\pi\over6}$ $\epsilon$ $\left[0,\pi\right]$

II) $\cos y = {{-\sqrt3}\over 2}$
$\Rightarrow$ $\cos y = \cos\left({-\pi\over6}\right)$

Since, $\cos(-x) = \cos(x)$
$\Rightarrow$ $\cos y = \cos\left({\pi\over6}\right)$
$\Rightarrow$ $y={\pi\over6}$ $\epsilon$ $\left[0,\pi\right]$

The I) is my textbok solution, and

II) is what i came up with.

Is my answer (II) correct?

Answer 1

$\cos\left(-\frac{\pi}{6}\right) = \cos\left(\frac{\pi}{6}\right)$

=$\frac{\sqrt{3}}{2}$

So your answer is incorrect.

Solution -

As we know cos function is negative in 2nd and 3rd quadrant

$-\frac{\sqrt3}{2} = \cos\left(π - \frac{π}{6}\right)$

or $\cos\left(π + \frac{π}{6}\right)$

as $\cos^{-1}$x range is in [ 0 , π ] and $\cos\left(π + \frac{π}{6}\right)$ ruled out.

So we have,

$-\frac{\sqrt3}{2} = \cos\left(π - \frac{π}{6}\right)$

I think now you can solve it further.

Answer 2

$-\frac{\sqrt{3}}{2}\neq\cos\left(-\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}$.

So your answer is incorrect.