Terrie Lang

2022-01-01

The general solution of $\left|\mathrm{sin}x\right|=\mathrm{cos}x$ is -

(A)$2n\pi +\frac{\pi}{4},n\in I$

(B)$2n\pi \pm \frac{\pi}{4},n\in I$

(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.

(A)

(B)

(C)

(D) None of these

So what I did was - I made a case for when

Melissa Moore

Beginner2022-01-02Added 32 answers

If

As x will lie in the 4th quadrant,

Jeffery Autrey

Beginner2022-01-03Added 35 answers

Using Weierstrass substitution with $t=\mathrm{tan}\frac{x}{2}$

$1-{t}^{2}=2\left|t\right|$

For real t,$t}^{2}={\left|t\right|}^{2$

$\Rightarrow {\left|t\right|}^{2}-2\left|t\right|-1=0$

$\Rightarrow \left|t\right|=1\pm \sqrt{2}$

As$\left|t\right|\ge 0,\left|t\right|=\sqrt{2}+1=\mathrm{csc}\frac{\pi}{4}+\mathrm{cot}\frac{\pi}{4}=\dots =\mathrm{cot}\frac{\pi}{8}=\mathrm{tan}(\frac{\pi}{2}-\frac{\pi}{8})$

$\iff {\mathrm{tan}}^{2}\frac{x}{2}={\mathrm{tan}}^{2}(\frac{\pi}{2}-\frac{\pi}{8})$

$\Rightarrow {x}^{2}=n\pi \pm (\frac{\pi}{2}-\frac{\pi}{8})$

For real t,

As

Vasquez

Expert2022-01-08Added 669 answers

We will solve

Here, we split into cases of

Checking these solutions in the original equation

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.

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