how do you find the $x$ value for $-\sin x+\cos x=0$

Question

Find the sationary points of the curve and their nature for the equation $y=e^x\cos x$ for $0\le x\le\pi/2$.

I derived it and got $e^x(-\sin x+\cos x)=0$.

$e^x$ has no solution but I don't know how to find the $x$ such that $-\sin x+\cos x=0$

Answer 1

Since you are asking for $0\le x \le \pi/2$ it's quite trivial that $x = \pi/4$ is the only solution of $\sin x = \cos x$

Edit:

Reasoning in $0\le x \le \pi/2$: notice that $\cos x = 0$ only if $x =\pi/2$ and for $x = \pi/2$ we have $\sin \pi/2 = 1 \ne 0 = \cos \pi /2$.

So we can say for sure that $x = \pi/2$ is not a solution of the equation $\sin x = \cos x$.

Divide by $\cos(x)$ both sides of $\sin x = \cos x$ and you get $\tan x = 1$ which is satiesfied in $0\le x \le \pi/2$ only for $x = \pi/4$

Answer 2

Hint:

$$-\sin x+\cos x=0$$ If and only if

$$\tan x=1$$

Can you get it from here?

Answer 3

We can first simplyfy the equation to have; $$\sin{x}=\cos{x}$$ $$\implies\sin^2{x}=\cos^2{x}.$$

Adding $\cos^2{x}$ to both sides gives; $$\sin^2{x}+\cos^2{x}=\cos^2{x}+\cos^2{x}=2\cos^2{x}.$$

Note that $\sin^2{x}+\cos^2{x}=1,$ so we know that; $$2\cos^2{x}=1$$ $$\implies\cos^2{x}=\frac{1}{2}$$ $$\implies\cos{x}=±\frac{\sqrt{2}}{2}.$$

Easily, the only solutions to this are $x=45^\circ, 135^\circ, 225^\circ,$ and $315.$ But since we also need $\sin{x}=\cos{x},$ the only solutions are $x=45^\circ$ and $225^\circ,$ or $x=\frac{\pi}{4}$ and $\frac{5\pi}{4}.$ But since $0\leq x\leq\frac{\pi}{2},$ $x=\frac{\pi}{4}$ is the only solution.

Answer 4

For all $x$, $\cos(x+\pi/2)=-\sin(x)$ hence the equation is equivalent to $$\cos(x)+\cos(x+\pi/2)=0$$

But we also have the identity

$$\cos p+\cos q=2\cos\frac{p+q}{2}\cos\frac{p-q}2$$

So your equation is also equivalent to

$$2\cos(x+\frac{\pi}4)\cos\frac{\pi}4=0$$

That is, $\cos(x+\frac\pi4)=0$.

But $\cos t=0$ iff $t=\frac{\pi}{2}+k\pi$, $k\in\Bbb Z$. Therefore, your equation is also equivalent to

$$x=\frac\pi4+k\pi$$

And the only value of $x$ in $[0,\pi/2]$ occurs for $k=0$.

Answer 5

Graphically we see immediately that $x=\frac {\pi}{4}$. But how about deriving this analytically?

Consider $\cos(2x)$ and use the formula for the double argument $\cos(2x) = \cos(x)^2 - \sin(x)^2$

Hence if $\sin(x) = \cos(x)$ we find that

$$\cos(2x)=0$$

Since $\cos(\alpha)$ vanishes if and only if the argument is of the form $\alpha=\pi(\frac{1}{2}+k), k=0,\pm1,\pm 2, ...$ $ we have

$$x = \frac{\pi}{2}(\frac{1}{2}+k)|_{k\to 0}\to \frac{\pi}{4}$$

in the requested range.

We can also find the value of $a=\sin(x)=\cos(x)\gt0$: from the relation $\sin(x)^2+\cos(x)^2=1$ we get $2 a^2 = 1$ and hence $a=\frac{1}{\sqrt{2}}$.