Find, in radians the general solution of \cos 3x = \sin 5

elvishwitchxyp 2022-01-02 Answered
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:
\(\displaystyle{3}{x}={2}{n}\pi-{\frac{{\pi}}{{{2}}}}+{5}{x}\)
\(\displaystyle{2}{x}={\frac{{\pi}}{{{2}}}}-{2}{n}\pi\)
\(\displaystyle{x}={\frac{{\pi}}{{{4}}}}-{n}\pi={\frac{{\pi}}{{{4}}}}{\left({1}−{4}{n}\right)}\)

Want to know more about Trigonometry?

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

censoratojk
Answered 2022-01-03 Author has 4897 answers
\(\displaystyle{\sin{{5}}}{x}={\cos{{\left({\frac{{\pi}}{{{2}}}}−{5}{x}\right)}}}={\cos{{3}}}{x}\)
\(\displaystyle{3}{x}={\frac{{\pi}}{{{2}}}}−{5}{x}+{2}{k}\pi\)
\(\displaystyle{x}={\frac{{\pi}}{{{16}}}}+{\frac{{{k}\pi}}{{{4}}}}={\frac{{\pi}}{{{16}}}}{\left({1}+{4}{k}\right)}\)
or
\(\displaystyle{3}{x}=−{\left({\frac{{\pi}}{{{2}}}}−{5}{x}\right)}+{2}{k}\pi\)
\(\displaystyle{x}={\frac{{\pi}}{{{4}}}}−{k}\pi\)
\(\displaystyle{x}={\frac{{\pi}}{{{4}}}}+{k}\pi={\frac{{\pi}}{{{4}}}}{\left({1}+{4}{k}\right)}\)
where \(\displaystyle{k}\in{Z}\)
writing \(\displaystyle-{k}\pi\ \text{ or }\ {k}\pi\) does not change the solution set. Because −k is the opposite of k in integers.
Not exactly what you’re looking for?
Ask My Question
0
 
Mary Nicholson
Answered 2022-01-04 Author has 3056 answers
If you write m in place of n, you reached at \(\displaystyle{\frac{{\pi{\left({1}−{4}{m}\right)}}}{{{4}}}}\)
We
\(\displaystyle{\frac{{\pi{\left({1}−{4}{m}\right)}}}{{{4}}}}={\frac{{\pi{\left({1}+{4}{n}\right)}}}{{{4}}}}\Leftrightarrow{m}=-{n}\)
In our case m is any integer \(\displaystyle\Leftrightarrow{n}=−{m}\) also belong to the same infinite set of integers
In their case n is so.
0
Vasquez
Answered 2022-01-09 Author has 8850 answers

No, the two are equivalent. In particular, if m = −n, then
\(\frac{\pi}{2}(1−4m)=\frac{\pi}{2}(4n+1),\)
so all that's really happened is tha tyou've listed the solutions in a different order.

0

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Relevant Questions

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
asked 2022-01-01
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.
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 2021-12-30
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 \(\displaystyle{x}={2}{n}\pi\pm{\frac{{\pi}}{{{3}}}}\)
asked 2021-08-19
If \(\displaystyle{\sin{{x}}}={\frac{{{1}}}{{{3}}}},\) x in quadrant 1, then find \(\displaystyle{\sin{{\left({2}{x}\right)}}},{\cos{{\left({2}{x}\right)}}},{\tan{{\left({2}{x}\right)}}}\)
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.
asked 2021-08-16
How do you solve \(\displaystyle{\sin{{x}}}{\cos{{x}}}={\frac{{{1}}}{{{2}}}}\) for x in the interval \(\displaystyle{\left[{0},{2}\pi\right)}\)?
...