crealolobk

2021-12-31

Find the sationary points of the curve and their nature for the equation $y={e}^{x}\mathrm{cos}x$ for $0\le x\le \frac{\pi }{2}$
I derived it and got ${e}^{x}\left(-\mathrm{sin}x+\mathrm{cos}x\right)=0$
${e}^{x}$ has no solution but I don't know how to find the x such that $-\mathrm{sin}x+\mathrm{cos}x=0$

boronganfh

Hint:
$-\mathrm{sin}x+\mathrm{cos}x=0$
If and only if
$\mathrm{tan}x=1$

aquariump9

Since you are asking for $0\le x\le \frac{\pi }{2}$ it's quite trivial that $x=\frac{\pi }{4}$ is the only solution of sinx=cosx
Edit:
Reasoning in $0\le x\le \frac{\pi }{2}$: notice that $\mathrm{cos}x=0$ only if $x=\frac{\pi }{2}$ and for $x=\frac{\pi }{2}$ we have $\mathrm{sin}\frac{\pi }{2}=1\ne 0=\mathrm{cos}\frac{\pi }{2}$
So we can say for sure that $x=\frac{\pi }{2}$ is not a solution of the equation $\mathrm{sin}x=\mathrm{cos}x$
Divide by $\mathrm{cos}\left(x\right)$ both sides of $\mathrm{sin}x=\mathrm{cos}x$ and you get $\mathrm{tan}x=1$ which is satiesfied in $0\le x\le \frac{\pi }{2}$ only for $x=\frac{\pi }{4}$

Vasquez

For all x, $\mathrm{cos}\left(x+\frac{\pi }{2}\right)=-\mathrm{sin}\left(x\right)$ hence the equation is equivalent to
$\mathrm{cos}\left(x\right)+\mathrm{cos}\left(x+\frac{\pi }{2}\right)=0$
But we also have the identity
$\mathrm{cos}p+\mathrm{cos}q=2\mathrm{cos}\frac{p+q}{2}\mathrm{cos}\frac{p-q}{2}$
So your equation is also equivalent to
$2\mathrm{cos}\left(x+\frac{\pi }{4}\right)\mathrm{cos}\frac{\pi }{4}=0$
That is, $\mathrm{cos}\left(x+\frac{\pi }{4}\right)=0$
But $\mathrm{cos}t=0$ if Therefore, your equation is also equivalent to
$x=\frac{\pi }{4}+k\pi$
And the only value of x in $\left[0,\frac{\pi }{2}\right]$ occurs for k=0.

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