Evaluate coordinate of turning point : $y={1-\sin x\over 1+\cos x}$

Question

How do I find the coordinate of the turning of $(1)$ $$y={1-\sin x\over 1+\cos x}={u\over v}\tag1$$

Using the Quotient rule: $${dy\over dx}={vu^{'}-v^{'}u\over v^2}$$ $u=1-\sin x$, $u^{'}=-\cos x$

$v=1+\cos x$, $v^{'}=-\sin x$

$${dy\over dx}={-\cos x(1+\cos x)+(1-\sin x)\sin x\over (1+\cos x)^2}$$ $$={-1-\cos x+\sin x\over (1+\cos x)^2}$$

Turning point happens ${dy\over dx}=0$.

$$1+\cos x=\sin x\tag2$$

How do I solve for $x$?

@King tut: $$\sqrt{2}\sin(x-{\pi\over 4})=1$$

$$x={\pi\over 2}$$

$$y={1-\pi\over 1+\pi}$$

Turning point $({\pi\over 2},{1-\pi\over 1+\pi})$

Answer 1

Hint for the trigonometric equation

$$1+\cos x=\sin x$$ $$\cos(0)+\cos x=\sin x$$ $$2\cos(x/2)\cos(x/2)=\sin x$$ $$2\cos^2(x/2)=\sin x$$ $$2\cos^2(x/2)=2\sin(x/2)\cos(x/2)$$ $$\cos(x/2)(\cos(x/2)-\sin(x/2))=0$$ $$ \implies \begin{cases} \cos(x/2)=0 \\ \cos(x/2)=\sin(x/2) \end{cases} $$ $$....$$

Answer 2

$y={1-\sin x\over 1+\cos x} = \frac{1}{1+\cos{x}}-\frac{\sin{x}}{1+\cos{x}} = \frac{1}{2} \left(1+\tan^2{\frac{x}{2}} \right) - \tan{\frac{x}{2}}$

$y' = \frac{1}{2}\tan{\frac{x}{2}}(1+\tan^2{\frac{x}{2}}) - \frac{1}{2}(1+\tan^2{\frac{x}{2}}) \stackrel{!}{=} 0 \Rightarrow \tan{\frac{x}{2}} = 1 \rightarrow x_t = \frac{\pi}{2}+2k\pi \; k \in \mathbb{Z}$

$y''(x) = \left( \frac{1}{2}(1+\tan^2{\frac{x}{2}})(\tan{\frac{x}{2}}-1) \right)' = \frac{1}{2}(\ldots)(\tan{\frac{x}{2}}-1) + \frac{1}{2}(1+\tan^2{\frac{x}{2}})\frac{1}{2}(1+\tan^2{\frac{x}{2}})$

$\Rightarrow y''(x_t) \gt 0$