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Question: If the equation \(\cos \left( 3x \right)+\sin \left( 2x-\dfrac{7\pi }{6} \right)=-2\) , then x is eq...

If the equation cos⁡(3x)+sin⁡(2x−7π6)=−2\cos \left( 3x \right)+\sin \left( 2x-\dfrac{7\pi }{6} \right)=-2 , then x is equal to k∈Zk\in Z.
A. π3(6k+1)\dfrac{\pi }{3}\left( 6k+1 \right)
B. π3(6k−1)\dfrac{\pi }{3}\left( 6k-1 \right)
C. π3(2k+1)\dfrac{\pi }{3}\left( 2k+1 \right)
D. None of these

Explanation

Solution

To solve this question, we should know the range of cos⁡x\cos x and sin⁡x\sin x. Both cos⁡x\cos x and sin⁡x\sin xvary between [−1,1]\left[ -1,1 \right]. If we observe the equation, the sum of the terms is given as -2 which is possible only when each of them is -1. So, cos⁡3x=−1\cos 3x=-1, sin⁡(2x−7π6)=−1\sin \left( 2x-\dfrac{7\pi }{6} \right)=-1. For cos⁡3x=−1\cos 3x=-1, the general solution of 3x is (2n+1)π, n∈Z\left( 2n+1 \right)\pi ,\text{ }n\in Z. Similarly, for sin⁡(2x−7π6)=−1\sin \left( 2x-\dfrac{7\pi }{6} \right)=-1, the general solution for (2x−7π6) is (4n−1)π2,n∈Z\left( 2x-\dfrac{7\pi }{6} \right)\text{ is }\left( 4n-1 \right)\dfrac{\pi }{2},n\in Z, using these solutions, we will write the solutions of x individually and the common solutions are the solutions for the required equation.

Complete step by step answer:
We know that the range of the function cos⁡x\cos x is [−1,1]\left[ -1,1 \right] and for the function sin⁡x\sin x is [−1,1]\left[ -1,1 \right].
In the question, we are given the equation cos⁡(3x)+sin⁡(2x−7π6)=−2\cos \left( 3x \right)+\sin \left( 2x-\dfrac{7\pi }{6} \right)=-2. As mentioned above, the minimum value of the two terms is -1. As the sum is equal to -2, we can write that each term is equal to -1.
cos⁡3x=−1 sin⁡(2x−7π6)=−1 \begin{aligned} & \cos 3x=-1 \\\ & \sin \left( 2x-\dfrac{7\pi }{6} \right)=-1 \\\ \end{aligned}
Let us consider cos⁡3x=−1\cos 3x=-1.
The function cos⁡y=−1\cos y=-1 for y=(2n+1)π,n∈Zy=\left( 2n+1 \right)\pi ,n\in Z
Here y = 3x,
3x=(2n+1)π x=(2n+1)π3 x=π3,3π3,5π3,7π3.......→(1) \begin{aligned} & 3x=\left( 2n+1 \right)\pi \\\ & x=\left( 2n+1 \right)\dfrac{\pi }{3} \\\ & x=\dfrac{\pi }{3},\dfrac{3\pi }{3},\dfrac{5\pi }{3},\dfrac{7\pi }{3}.......\to \left( 1 \right) \\\ \end{aligned}

Let us consider sin⁡(2x−7π6)=−1\sin \left( 2x-\dfrac{7\pi }{6} \right)=-1.
The function sin⁡y=−1\sin y=-1 for y=(4n−1)π2,n∈Zy=\left( 4n-1 \right)\dfrac{\pi }{2},n\in Z
Here y=2x−7π6y=2x-\dfrac{7\pi }{6},
2x−7π6=(4n−1)π2 2x=(4n−1)π2+7π6 2x=2nπ+2π3 x=nπ+π3 x=π3,4π3,7π3,10π3.......→(2) \begin{aligned} & 2x-\dfrac{7\pi }{6}=\left( 4n-1 \right)\dfrac{\pi }{2} \\\ & 2x=\left( 4n-1 \right)\dfrac{\pi }{2}+\dfrac{7\pi }{6} \\\ & 2x=2n\pi +\dfrac{2\pi }{3} \\\ & x=n\pi +\dfrac{\pi }{3} \\\ & x=\dfrac{\pi }{3},\dfrac{4\pi }{3},\dfrac{7\pi }{3},\dfrac{10\pi }{3}.......\to \left( 2 \right) \\\ \end{aligned}
From the two equations-1, 2 we should take the common solutions. The common solutions are
x=π3,7π3,13π3.....x=\dfrac{\pi }{3},\dfrac{7\pi }{3},\dfrac{13\pi }{3}.....
The general solution can be written as
x=(6k+1)π3,k∈Zx=\dfrac{\left( 6k+1 \right)\pi }{3},k\in Z.
∴x=(6k+1)π3\therefore x=\dfrac{\left( 6k+1 \right)\pi }{3}.

So, the correct answer is “Option A”.

Note: Some students tend to alter the L.H.S of the equation by writing different transformation formulae of sine and cosine functions. They lead to complex calculations which can be solved but will waste a lot of time. The main trick in this question is to know that both the terms in the equation are equal to -1. In these types of questions, when the sum of ‘n’ sine or cosine functions is equal to n or –n, every value of them is equal to 1 or -1 respectively.