The general solution of |\sin x|=\cos x is - (A)

Terrie Lang 2022-01-01 Answered
The general solution of \(\displaystyle{\left|{\sin{{x}}}\right|}={\cos{{x}}}\) is -
(A) \(\displaystyle{2}{n}\pi+{\frac{{\pi}}{{{4}}}},{n}\in{I}\)
(B) \(\displaystyle{2}{n}\pi\pm{\frac{{\pi}}{{{4}}}},{n}\in{I}\)
(C) \(\displaystyle{n}\pi+{\frac{{\pi}}{{{4}}}},{n}\in{I}\)
(D) None of these
So what I did was - I made a case for when \(\displaystyle{\sin{{x}}}\) is greater than 0 and equated it to \(\displaystyle{\cos{{x}}}\) to get \(\displaystyle{\tan{{x}}}={1}\) which implies \(\displaystyle{x}={\frac{{\pi}}{{{4}}}}\). The other case was when \(\displaystyle{\cos{{x}}}=−{\sin{{x}}}\). Here, \(\displaystyle{x}={\frac{{{3}\pi}}{{{4}}}}\). I don't understand how to proceed from here.

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Expert Answer

Melissa Moore
Answered 2022-01-02 Author has 3929 answers
\(\displaystyle{\cos{{x}}}={\left|{\sin{{x}}}\right|}\geq{0}\)
If \(\displaystyle{\sin{{x}}}{ < }{0},{\cos{{x}}}=−{\sin{{x}}}\Leftrightarrow{\tan{{x}}}=-{1}\)
As x will lie in the 4th quadrant, \(\displaystyle{x}={2}{n}\pi−{\frac{{\pi}}{{{4}}}}\)
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Jeffery Autrey
Answered 2022-01-03 Author has 3319 answers
Using Weierstrass substitution with \(\displaystyle{t}={\tan{{\frac{{{x}}}{{{2}}}}}}\)
\(\displaystyle{1}-{t}^{{2}}={2}{\left|{t}\right|}\)
For real t, \(\displaystyle{t}^{{2}}={\left|{t}\right|}^{{2}}\)
\(\displaystyle\Rightarrow{\left|{t}\right|}^{{2}}-{2}{\left|{t}\right|}−{1}={0}\)
\(\displaystyle\Rightarrow{\left|{t}\right|}={1}\pm\sqrt{{2}}\)
As \(\displaystyle{\left|{t}\right|}\geq{0},{\left|{t}\right|}=\sqrt{{2}}+{1}={\csc{{\frac{{\pi}}{{{4}}}}}}+{\cot{{\frac{{\pi}}{{{4}}}}}}=\ldots={\cot{{\frac{{\pi}}{{{8}}}}}}={\tan{{\left({\frac{{\pi}}{{{2}}}}-{\frac{{\pi}}{{{8}}}}\right)}}}\)
\(\displaystyle\Leftrightarrow{{\tan}^{{2}}{\frac{{{x}}}{{{2}}}}}={{\tan}^{{2}}{\left({\frac{{\pi}}{{{2}}}}−{\frac{{\pi}}{{{8}}}}\right)}}\)
\(\displaystyle\Rightarrow{x}^{{2}}={n}\pi\pm{\left({\frac{{\pi}}{{{2}}}}−{\frac{{\pi}}{{{8}}}}\right)}\)
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Vasquez
Answered 2022-01-08 Author has 8850 answers

We will solve \(\sin^2(x)=\cos^2(x)\), which is obtained upon squaring the equation (so we do not have to deal with sign issues).
\(\begin{array}{}\sin^2(x)=\cos^2(x) \\\cos^2(x)−\sin^2(x)=0 \\\cos(2x)=0 \\x=\frac{\pi}{4}+\pi n, \frac{3\pi}{4}+\pi n \\x=\frac{\pi}{4}+2\pi n,\frac{5\pi}{4}+2\pi n,\frac{3\pi}{4}+2\pi n,\frac{7\pi}{4}+2\pi n \end{array}\)
Here, we split into cases of \(2\pi n\) periodicity to take advantage of the properties of \(\sin\)and \(\cos\)
Checking these solutions in the original equation \(|\sin(x)|=\cos(x)\), 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) \(\pm \frac{\pi}{4}+2\pi n\)
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.

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