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Question: Solve the equation \(\cos 5x = \sin x\)?...

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

Explanation

Solution

Here we will use two formulas two solve this question, one among them is sin⁡x=cos⁡(π2−x)\sin x = \cos \left( {\dfrac{\pi }{2} - x} \right) and the other one is, if cos⁡θ=cos⁡α ⇒   θ=α±2nπ, where  n∈Z\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\cos 5x = \sin x.
We know that
sin⁡x=cos⁡(π2−x)\sin x = \cos \left( {\dfrac{\pi }{2} - x} \right)
Here,
cos5x=sinx  , where  0⩽x⩽360∘cos5x = sinx\;,{\text{ }}where\;0 \leqslant x \leqslant {360^ \circ }
⇒cos⁡5x=cos⁡(π2−x)\Rightarrow \cos 5x = \cos \left( {\dfrac{\pi }{2} - x} \right)
Now, using the property, ifcos⁡θ=cos⁡α ⇒   θ=α±2nπ, where  n∈Z\cos \theta = \cos \alpha \, \Rightarrow \,\,\,\theta = \alpha \pm 2n\pi ,\,where\,\,n \in Z
⇒5x=2nπ±(π2−x),n∈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.
⇒5x=2nπ+π2−x  or   5x=2nπ−π2+x,n∈Z\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=2nπ+π2−x,n∈Z5x = 2n\pi + \dfrac{\pi }{2} - x,n \in Z
On transposing x, we get
⇒5x+x=2kπ+π2,n∈Z\Rightarrow 5x + x = 2k\pi + \dfrac{\pi }{2},n \in Z
⇒6x=(4k+1)π2,n∈Z\Rightarrow 6x = (4k + 1)\dfrac{\pi }{2},n \in Z
⇒x=(4k+1)π12,k∈Z\Rightarrow x = (4k + 1)\dfrac{\pi }{{12}},k \in Z
n=0⇒x=π12andπ12∈[0,2π]n = 0 \Rightarrow x = \dfrac{\pi }{{12}}and\dfrac{\pi }{{12}} \in [0,2\pi ]
n=1⇒x=5π12and5π12∈[0,2π]n = 1 \Rightarrow x = \dfrac{{5\pi }}{{12}}and\dfrac{{5\pi }}{{12}} \in [0,2\pi ]
n=2⇒x=9π12and9π12∈[0,2π]n = 2 \Rightarrow x = \dfrac{{9\pi }}{{12}}and\dfrac{{9\pi }}{{12}} \in [0,2\pi ]
n=3⇒x=13π12and13π12∈[0,2π]n = 3 \Rightarrow x = \dfrac{{13\pi }}{{12}}and\dfrac{{13\pi }}{{12}} \in [0,2\pi ] and so on…
Now,
Case II:
5x=2nπ−π2+x,n∈Z5x = 2n\pi - \dfrac{\pi }{2} + x,n \in Z
⇒5x−x=2nπ−π2,n∈Z\Rightarrow 5x - x = 2n\pi - \dfrac{\pi }{2},n \in Z
⇒4x=2nπ−π2,n∈Z\Rightarrow 4x = 2n\pi - \dfrac{\pi }{2},n \in Z
⇒4x=(4n−1)π2,n∈Z\Rightarrow 4x = (4n - 1)\dfrac{\pi }{2},n \in Z
⇒x=(4n−1)π8,n∈Z\Rightarrow x = (4n - 1)\dfrac{\pi }{8},n \in Z
n=0⇒x=−π8and−π8∉[0,2π]n = 0 \Rightarrow x = - \dfrac{\pi }{8}and - \dfrac{\pi }{8} \notin [0,2\pi ]
n=1⇒x=3π8and3π8∈[0,2π]n = 1 \Rightarrow x = \dfrac{{3\pi }}{8}and\dfrac{{3\pi }}{8} \in [0,2\pi ]
n=2⇒x=7π8and7π8∈[0,2π]n = 2 \Rightarrow x = \dfrac{{7\pi }}{8}and\dfrac{{7\pi }}{8} \in [0,2\pi ]
n=3⇒x=11π8and11π8∈[0,2π]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=π12,5π12,9π12,13π12,17π12,21π12,3π8,7π8,11π8,15π8.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.