[tex]\bf \textit{let's recall that }sin^2(\theta)+cos^2(\theta)=1
\\\\\\
sin^2(\theta)=1-cos^2(\theta)\implies sin(\theta )=\sqrt{1-cos^2(\theta)}\\\\
-------------------------------\\\\[/tex]
[tex]\bf y=sin(x)cos(x)\impliedby \textit{now, using the product-rule}
\\\\\\
\cfrac{dy}{dx}=cos(x)cos(x)-sin(x)cos(x)\implies 0=cos(x)[cos(x)-sin(x)]\\\\
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0=cos(x)\implies \boxed{\measuredangle x=
\begin{cases}
\frac{\pi }{2}\\\\
\frac{3\pi }{2}
\end{cases}}\\\\
-------------------------------\\\\[/tex]
[tex]\bf 0=cos(x)-sin(x)\implies sin(x)=cos(x)\\\\
thus\qquad \sqrt{1-cos^2(x)}=cos(x)\implies 1-cos^2(x)=cos^2(x)
\\\\\\
1=2cos^2(x)\implies \cfrac{1}{2}=cos^2(x)\implies \pm\sqrt{\cfrac{1}{2}}=cos(x)
\\\\\\
\pm\cfrac{1}{\sqrt{2}}=cos(x)\implies \pm\cfrac{\sqrt{2}}{2}=cos(x)\implies \boxed{\measuredangle x=
\begin{cases}
\frac{\pi }{4}\\\\
\frac{3\pi }{4}\\\\
\frac{5\pi }{4}\\\\
\frac{7\pi }{4}
\end{cases}}[/tex]
