 Terrie Lang

2022-01-01

The general solution of $|\mathrm{sin}x|=\mathrm{cos}x$ is -
(A) $2n\pi +\frac{\pi }{4},n\in I$
(B) $2n\pi ±\frac{\pi }{4},n\in I$
(C) $n\pi +\frac{\pi }{4},n\in I$
(D) None of these
So what I did was - I made a case for when $\mathrm{sin}x$ is greater than 0 and equated it to $\mathrm{cos}x$ to get $\mathrm{tan}x=1$ which implies $x=\frac{\pi }{4}$. The other case was when $\mathrm{cos}x=-\mathrm{sin}x$. Here, $x=\frac{3\pi }{4}$. I don't understand how to proceed from here. Melissa Moore

Expert

$\mathrm{cos}x=|\mathrm{sin}x|\ge 0$
If $\mathrm{sin}x<0,\mathrm{cos}x=-\mathrm{sin}x⇔\mathrm{tan}x=-1$
As x will lie in the 4th quadrant, $x=2n\pi -\frac{\pi }{4}$ Jeffery Autrey

Expert

Using Weierstrass substitution with $t=\mathrm{tan}\frac{x}{2}$
$1-{t}^{2}=2|t|$
For real t, ${t}^{2}={|t|}^{2}$
$⇒{|t|}^{2}-2|t|-1=0$
$⇒|t|=1±\sqrt{2}$
As $|t|\ge 0,|t|=\sqrt{2}+1=\mathrm{csc}\frac{\pi }{4}+\mathrm{cot}\frac{\pi }{4}=\dots =\mathrm{cot}\frac{\pi }{8}=\mathrm{tan}\left(\frac{\pi }{2}-\frac{\pi }{8}\right)$
$⇔{\mathrm{tan}}^{2}\frac{x}{2}={\mathrm{tan}}^{2}\left(\frac{\pi }{2}-\frac{\pi }{8}\right)$
$⇒{x}^{2}=n\pi ±\left(\frac{\pi }{2}-\frac{\pi }{8}\right)$ Vasquez

Expert

We will solve ${\mathrm{sin}}^{2}\left(x\right)={\mathrm{cos}}^{2}\left(x\right)$, which is obtained upon squaring the equation (so we do not have to deal with sign issues).
$\begin{array}{}{\mathrm{sin}}^{2}\left(x\right)={\mathrm{cos}}^{2}\left(x\right)\\ {\mathrm{cos}}^{2}\left(x\right)-{\mathrm{sin}}^{2}\left(x\right)=0\\ \mathrm{cos}\left(2x\right)=0\\ x=\frac{\pi }{4}+\pi n,\frac{3\pi }{4}+\pi n\\ x=\frac{\pi }{4}+2\pi n,\frac{5\pi }{4}+2\pi n,\frac{3\pi }{4}+2\pi n,\frac{7\pi }{4}+2\pi n\end{array}$
Here, we split into cases of $2\pi n$ periodicity to take advantage of the properties of $\mathrm{sin}$and $\mathrm{cos}$
Checking these solutions in the original equation $|\mathrm{sin}\left(x\right)|=\mathrm{cos}\left(x\right)$, we find that only $\frac{\pi }{4}+2\pi n$ and $\frac{7\pi }{4}+2\pi n$ work. $\frac{7\pi }{4}+2\pi n$ is equivalent to $-\frac{\pi }{4}+2\pi n$, so the answer is (B) $±\frac{\pi }{4}+2\pi n$
You can do a sanity check by drawing a quick graph and seeing if the functions intersect in the general region of the conjectured solution.

Do you have a similar question?