If \(\displaystyle{\sin{{x}}}{ < }{0},{\cos{{x}}}=−{\sin{{x}}}\Leftrightarrow{\tan{{x}}}=-{1}\)

As x will lie in the 4th quadrant, \(\displaystyle{x}={2}{n}\pi−{\frac{{\pi}}{{{4}}}}\)

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 \(\displaystyle{\left|{t}\right|}\geq{0},{\left|{t}\right|}=\sqrt{{2}}+{1}={\csc{{\frac{{\pi}}{{{4}}}}}}+{\cot{{\frac{{\pi}}{{{4}}}}}}=\ldots={\cot{{\frac{{\pi}}{{{8}}}}}}={\tan{{\left({\frac{{\pi}}{{{2}}}}-{\frac{{\pi}}{{{8}}}}\right)}}}\)

\(\displaystyle\Leftrightarrow{{\tan}^{{2}}{\frac{{{x}}}{{{2}}}}}={{\tan}^{{2}}{\left({\frac{{\pi}}{{{2}}}}−{\frac{{\pi}}{{{8}}}}\right)}}\)

\(\displaystyle\Rightarrow{x}^{{2}}={n}\pi\pm{\left({\frac{{\pi}}{{{2}}}}−{\frac{{\pi}}{{{8}}}}\right)}\)

\(\displaystyle{1}-{t}^{{2}}={2}{\left|{t}\right|}\)

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 \(\displaystyle{\left|{t}\right|}\geq{0},{\left|{t}\right|}=\sqrt{{2}}+{1}={\csc{{\frac{{\pi}}{{{4}}}}}}+{\cot{{\frac{{\pi}}{{{4}}}}}}=\ldots={\cot{{\frac{{\pi}}{{{8}}}}}}={\tan{{\left({\frac{{\pi}}{{{2}}}}-{\frac{{\pi}}{{{8}}}}\right)}}}\)

\(\displaystyle\Leftrightarrow{{\tan}^{{2}}{\frac{{{x}}}{{{2}}}}}={{\tan}^{{2}}{\left({\frac{{\pi}}{{{2}}}}−{\frac{{\pi}}{{{8}}}}\right)}}\)

\(\displaystyle\Rightarrow{x}^{{2}}={n}\pi\pm{\left({\frac{{\pi}}{{{2}}}}−{\frac{{\pi}}{{{8}}}}\right)}\)

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.

asked 2021-12-31

What is general solution of \(\displaystyle{\frac{{{\cos{{5}}}{x}{\cos{{3}}}{x}-{\sin{{3}}}{x}{\sin{{x}}}}}{{{\cos{{2}}}{x}}}}={1}\)

1) \(\displaystyle{\frac{{{k}\pi}}{{{3}}}}\)

2) \(\displaystyle{\frac{{{k}\pi}}{{{2}}}}\)

3) \(\displaystyle{\frac{{{2}{k}\pi}}{{{5}}}}\)

4) \(\displaystyle{\frac{{{2}{k}\pi}}{{{3}}}}\)

The numerator of the fraction is \(\displaystyle{\cos{{\left({5}{x}+{3}{x}\right)}}}\). so I should find general solution of \(\displaystyle{\cos{{8}}}{x}={\cos{{2}}}{x}\). I'm not sure how to do it, I can write \(\displaystyle{\cos{{8}}}{x}\) in term of \(\displaystyle{\cos{{2}}}{x}\):

\(\displaystyle{\cos{{8}}}{x}={2}{{\cos}^{{2}}{4}}{x}−{1}={2}{\left({2}{{\cos}^{{2}}{2}}{x}−{1}\right)}^{{2}}−{1}\)

After substituting it in the equation and using \(\displaystyle{\cos{{2}}}{x}={t}\) we have degree four equation

1) \(\displaystyle{\frac{{{k}\pi}}{{{3}}}}\)

2) \(\displaystyle{\frac{{{k}\pi}}{{{2}}}}\)

3) \(\displaystyle{\frac{{{2}{k}\pi}}{{{5}}}}\)

4) \(\displaystyle{\frac{{{2}{k}\pi}}{{{3}}}}\)

The numerator of the fraction is \(\displaystyle{\cos{{\left({5}{x}+{3}{x}\right)}}}\). so I should find general solution of \(\displaystyle{\cos{{8}}}{x}={\cos{{2}}}{x}\). I'm not sure how to do it, I can write \(\displaystyle{\cos{{8}}}{x}\) in term of \(\displaystyle{\cos{{2}}}{x}\):

\(\displaystyle{\cos{{8}}}{x}={2}{{\cos}^{{2}}{4}}{x}−{1}={2}{\left({2}{{\cos}^{{2}}{2}}{x}−{1}\right)}^{{2}}−{1}\)

After substituting it in the equation and using \(\displaystyle{\cos{{2}}}{x}={t}\) we have degree four equation

asked 2022-01-02

Find, in radians the general solution of \(\displaystyle{\cos{{3}}}{x}={\sin{{5}}}{x}\)

I have said, \(\displaystyle{\sin{{5}}}{x}={\cos{{\left({\frac{{\pi}}{{{2}}}}−{5}{x}\right)}}}\)

so

\(\displaystyle{\cos{{3}}}{x}={\sin{{5}}}{x}\Rightarrow{3}{x}={2}{n}\pi\pm{\left({\frac{{\pi}}{{{2}}}}-{5}{x}\right)}\)

When I add \(\displaystyle{\left({\frac{{\pi}}{{{2}}}}−{5}{x}\right)}\ \text{ to }\ {2}{n}\pi\) I get the answer \(\displaystyle{x}={\frac{{\pi}}{{{16}}}}{\left({4}{n}+{1}\right)}\), which the book says is correct.

But when I subtract I get a different answer to the book. My working is as follows:

\(\displaystyle{3}{x}={2}{n}\pi-{\frac{{\pi}}{{{2}}}}+{5}{x}\)

\(\displaystyle{2}{x}={\frac{{\pi}}{{{2}}}}-{2}{n}\pi\)

\(\displaystyle{x}={\frac{{\pi}}{{{4}}}}-{n}\pi={\frac{{\pi}}{{{4}}}}{\left({1}−{4}{n}\right)}\)

I have said, \(\displaystyle{\sin{{5}}}{x}={\cos{{\left({\frac{{\pi}}{{{2}}}}−{5}{x}\right)}}}\)

so

\(\displaystyle{\cos{{3}}}{x}={\sin{{5}}}{x}\Rightarrow{3}{x}={2}{n}\pi\pm{\left({\frac{{\pi}}{{{2}}}}-{5}{x}\right)}\)

When I add \(\displaystyle{\left({\frac{{\pi}}{{{2}}}}−{5}{x}\right)}\ \text{ to }\ {2}{n}\pi\) I get the answer \(\displaystyle{x}={\frac{{\pi}}{{{16}}}}{\left({4}{n}+{1}\right)}\), which the book says is correct.

But when I subtract I get a different answer to the book. My working is as follows:

\(\displaystyle{3}{x}={2}{n}\pi-{\frac{{\pi}}{{{2}}}}+{5}{x}\)

\(\displaystyle{2}{x}={\frac{{\pi}}{{{2}}}}-{2}{n}\pi\)

\(\displaystyle{x}={\frac{{\pi}}{{{4}}}}-{n}\pi={\frac{{\pi}}{{{4}}}}{\left({1}−{4}{n}\right)}\)

asked 2021-12-30

Find the general solution of equation \(\displaystyle{\cot{{x}}}+{\tan{{x}}}={2}\)

My approach:

\(\displaystyle{\cot{{x}}}+{\tan{{x}}}={\frac{{{\left({1}+{{\tan}^{{2}}{x}}\right)}}}{{{\tan{{x}}}}}}={2}\)

\(\displaystyle{{\sec}^{{2}}{x}}+{\tan{{x}}}={2}\)

\(\displaystyle{\frac{{{1}}}{{{\left({\sin{{x}}}{\cos{{x}}}\right)}}}}={2}\)

\(\displaystyle{{\sin}^{{2}}{x}}+{{\cos}^{{2}}{x}}={2}{\sin{{x}}}{\cos{{x}}}\)

\(\displaystyle{\left({\sin{{x}}}−{\cos{{x}}}\right)}^{{2}}={0}\)

\(\displaystyle{\sin{{x}}}={\cos{{x}}}\)

So, \(\displaystyle{x}={n}\pi+{\left({\frac{{\pi}}{{{4}}}}\right)}\). This was my answer. But the answer given is \(\displaystyle{x}={2}{n}\pi\pm{\frac{{\pi}}{{{3}}}}\)

My approach:

\(\displaystyle{\cot{{x}}}+{\tan{{x}}}={\frac{{{\left({1}+{{\tan}^{{2}}{x}}\right)}}}{{{\tan{{x}}}}}}={2}\)

\(\displaystyle{{\sec}^{{2}}{x}}+{\tan{{x}}}={2}\)

\(\displaystyle{\frac{{{1}}}{{{\left({\sin{{x}}}{\cos{{x}}}\right)}}}}={2}\)

\(\displaystyle{{\sin}^{{2}}{x}}+{{\cos}^{{2}}{x}}={2}{\sin{{x}}}{\cos{{x}}}\)

\(\displaystyle{\left({\sin{{x}}}−{\cos{{x}}}\right)}^{{2}}={0}\)

\(\displaystyle{\sin{{x}}}={\cos{{x}}}\)

So, \(\displaystyle{x}={n}\pi+{\left({\frac{{\pi}}{{{4}}}}\right)}\). This was my answer. But the answer given is \(\displaystyle{x}={2}{n}\pi\pm{\frac{{\pi}}{{{3}}}}\)

asked 2021-08-16

If \(\displaystyle{\sin{{\left({3}{x}+{5}\right)}}}={\cos{{\left({2}{x}-{10}\right)}}}\) what is the value of x?

asked 2021-12-31

Solve: \(\displaystyle{3}{\sin{{2}}}{x}+{4}{\cos{{2}}}{x}−{2}{\cos{{x}}}+{6}{\sin{{x}}}-{6}={0}\)

My Try

\(\displaystyle{6}{\sin{{x}}}{\cos{{x}}}+{4}{\left({{\cos}^{{2}}{x}}−{{\sin}^{{2}}{x}}\right)}−{2}{\cos{{x}}}+{6}{\sin{{x}}}−{6}={0}\)

My Try

\(\displaystyle{6}{\sin{{x}}}{\cos{{x}}}+{4}{\left({{\cos}^{{2}}{x}}−{{\sin}^{{2}}{x}}\right)}−{2}{\cos{{x}}}+{6}{\sin{{x}}}−{6}={0}\)

asked 2021-12-31

How does \(\displaystyle{\frac{{{4}\pi{x}+{\cos{{\left({4}\pi{x}\right)}}}{\sin{{\left({4}\pi{x}\right)}}}}}{{{8}\pi}}}+{C}\) become \(\displaystyle{\frac{{{\sin{{\left({8}\pi{x}\right)}}}+{8}\pi{x}}}{{{16}\pi}}}+{C}\)?

asked 2021-12-31

Basically, write \(\displaystyle{\cos{{4}}}{x}\) as a polynomial in \(\displaystyle{\sin{{x}}}\).

I've tried the double angles theorem and \(\displaystyle{\cos{{2}}}{x}={{\cos}^{{2}}{x}}-{{\sin}^{{2}}{x}}\). I'm still having trouble right now though.

I've tried the double angles theorem and \(\displaystyle{\cos{{2}}}{x}={{\cos}^{{2}}{x}}-{{\sin}^{{2}}{x}}\). I'm still having trouble right now though.