Question

Solve the equation $\cos 5x = \sin x$?

Answer 1

Hint: Here we will use two formulas two solve this question, one among them is $\sin x = \cos \left( {\dfrac{\pi }{2} - x} \right)$ and the other one is, if $\cos \theta = \cos \alpha \, \Rightarrow \,\,\,\theta = \alpha \pm 2n\pi ,\,where\,\,n \in Z$.
Here we have to find the particular solution and there can be two cases, one with plus sign and one with minus sign.

Complete step-by-step answer:
In the above question, it is given that $\cos 5x = \sin x$.
We know that
$\sin x = \cos \left( {\dfrac{\pi }{2} - x} \right)$
Here,
\[cos5x = sinx\;,{\text{ }}where\;0 \leqslant x \leqslant {360^ \circ }\]
$ \Rightarrow \cos 5x = \cos \left( {\dfrac{\pi }{2} - x} \right)$
Now, using the property, if$\cos \theta = \cos \alpha \, \Rightarrow \,\,\,\theta = \alpha \pm 2n\pi ,\,where\,\,n \in Z$
\[ \Rightarrow 5x = 2n\pi \pm \left( {\dfrac{\pi }{2} - x} \right),n \in Z.\]
Now, there are two solutions. One with plus sign and second one with minus sign.
\[ \Rightarrow 5x = 2n\pi + \dfrac{\pi }{2} - x\,\,or\,\,\,5x = 2n\pi - \dfrac{\pi }{2} + x,n \in Z\]
Now there are two cases:
Case I:
\[5x = 2n\pi + \dfrac{\pi }{2} - x,n \in Z\]
On transposing x, we get
\[ \Rightarrow 5x + x = 2k\pi + \dfrac{\pi }{2},n \in Z\]
\[ \Rightarrow 6x = (4k + 1)\dfrac{\pi }{2},n \in Z\]
\[ \Rightarrow x = (4k + 1)\dfrac{\pi }{{12}},k \in Z\]
\[n = 0 \Rightarrow x = \dfrac{\pi }{{12}}and\dfrac{\pi }{{12}} \in [0,2\pi ]\]
\[n = 1 \Rightarrow x = \dfrac{{5\pi }}{{12}}and\dfrac{{5\pi }}{{12}} \in [0,2\pi ]\]
\[n = 2 \Rightarrow x = \dfrac{{9\pi }}{{12}}and\dfrac{{9\pi }}{{12}} \in [0,2\pi ]\]
\[n = 3 \Rightarrow x = \dfrac{{13\pi }}{{12}}and\dfrac{{13\pi }}{{12}} \in [0,2\pi ]\] and so on…
Now,
Case II:
\[5x = 2n\pi - \dfrac{\pi }{2} + x,n \in Z\]
\[ \Rightarrow 5x - x = 2n\pi - \dfrac{\pi }{2},n \in Z\]
\[ \Rightarrow 4x = 2n\pi - \dfrac{\pi }{2},n \in Z\]
\[ \Rightarrow 4x = (4n - 1)\dfrac{\pi }{2},n \in Z\]
\[ \Rightarrow x = (4n - 1)\dfrac{\pi }{8},n \in Z\]
\[n = 0 \Rightarrow x = - \dfrac{\pi }{8}and - \dfrac{\pi }{8} \notin [0,2\pi ]\]
\[n = 1 \Rightarrow x = \dfrac{{3\pi }}{8}and\dfrac{{3\pi }}{8} \in [0,2\pi ]\]
\[n = 2 \Rightarrow x = \dfrac{{7\pi }}{8}and\dfrac{{7\pi }}{8} \in [0,2\pi ]\]
\[n = 3 \Rightarrow x = \dfrac{{11\pi }}{8}and\dfrac{{11\pi }}{8} \in [0,2\pi ]\] and so on…
Therefore, the solution of the given equation is:
\[x = \dfrac{\pi }{{12}},\dfrac{{5\pi }}{{12}},\dfrac{{9\pi }}{{12}},\dfrac{{13\pi }}{{12}},\dfrac{{17\pi }}{{12}},\dfrac{{21\pi }}{{12}},\dfrac{{3\pi }}{8},\dfrac{{7\pi }}{8},\dfrac{{11\pi }}{8},\dfrac{{15\pi }}{8}.\]

Note: In this question, we can also convert the equation in terms of sine function instead of cosine function and after that the whole procedure is the same for the whole question. There are some solutions which are not acceptable for a particular value of n in the given domain.