Question

If $\cos 3x+\sin \left( 2x-\dfrac{7\pi }{6} \right)=-2$ then $x$ is equal to $\left( k\in z \right)$
(a) $\dfrac{\pi }{3}\left( 6k+1 \right)$
(b) $\dfrac{\pi }{3}\left( 6k-1 \right)$
(c) $\dfrac{\pi }{3}\left( 2k+1 \right)$
(d) None of these

Answer 1

We need to add one two times to the given equation to simplify it as $1+\cos 3x+1+\sin \left( 2x-\dfrac{7\pi }{6} \right)=0$. Then on adding and subtracting $\dfrac{\pi }{2}$ to the sine argument and by using the identity $\sin \left( \dfrac{\pi }{2}-\theta \right)=\cos \theta$ we will obtain the given equation as ${{\cos }^{2}}\dfrac{3x}{2}+{{\cos }^{2}}\left( \dfrac{5\pi }{6}-x \right)=0$, from which we will obtain $\cos \dfrac{3x}{2}=0$ and $\cos \left( \dfrac{5\pi }{6}-x \right)=0$ which can be solved by using the solution of the trigonometric equation $\cos x=0$ which is given by $x=\left( 2n+1 \right)\dfrac{\pi }{2}$.

Complete step by step solution:
The trigonometric equation given in the above question is written as
$\Rightarrow \cos 3x+\sin \left( 2x-\dfrac{7\pi }{6} \right)=-2$
Adding one both the sides, we get
$\begin{aligned} & \Rightarrow 1+\cos 3x+\sin \left( 2x-\dfrac{7\pi }{6} \right)=-2+1 \\\ & \Rightarrow 1+\cos 3x+\sin \left( 2x-\dfrac{7\pi }{6} \right)=-1 \\\ \end{aligned}$
Again, on adding one both the sides, we get
$\begin{aligned} & \Rightarrow 1+\cos 3x+1+\sin \left( 2x-\dfrac{7\pi }{6} \right)=-1+1 \\\ & \Rightarrow 1+\cos 3x+1+\sin \left( 2x-\dfrac{7\pi }{6} \right)=0......\left( i \right) \\\ \end{aligned}$
Now, we know the trigonometric identity given by
$\Rightarrow 1+\cos \theta ={{\cos }^{2}}\dfrac{\theta }{2}$
On substituting $\theta =3x$ in the above identity we get
$\Rightarrow 1+\cos 3x={{\cos }^{2}}\dfrac{3x}{2}$
Substituting the above equation in the equation (i) we get
$\Rightarrow {{\cos }^{2}}\dfrac{3x}{2}+1+\sin \left( 2x-\dfrac{7\pi }{6} \right)=0$
Adding and subtracting $\dfrac{\pi }{2}$ in the argument of the sine term in the above equation we get

& \Rightarrow {{\cos }^{2}}\dfrac{3x}{2}+1+\sin \left( 2x-\dfrac{7\pi }{6}+\dfrac{\pi }{2}-\dfrac{\pi }{2} \right)=0 \\\ & \Rightarrow {{\cos }^{2}}\dfrac{3x}{2}+1+\sin \left( \dfrac{\pi }{2}+2x-\dfrac{10\pi }{6} \right)=0 \\\ & \Rightarrow {{\cos }^{2}}\dfrac{3x}{2}+1+\sin \left( \dfrac{\pi }{2}+2x-\dfrac{5\pi }{3} \right)=0 \\\ & \Rightarrow {{\cos }^{2}}\dfrac{3x}{2}+1+\sin \left( \dfrac{\pi }{2}-\left( \dfrac{5\pi }{3}-2x \right) \right)=0 \\\ \end{aligned}$$ Now, using the trigonometric identity $$\sin \left( \dfrac{\pi }{2}-\theta \right)=\cos \theta $$, we can write the above equation as $$\Rightarrow {{\cos }^{2}}\dfrac{3x}{2}+1+\cos \left( \dfrac{5\pi }{3}-2x \right)=0$$ Now, we can again apply the trigonometric identity $$1+\cos \theta ={{\cos }^{2}}\dfrac{\theta }{2}$$ to write the above equation as $$\begin{aligned} & \Rightarrow {{\cos }^{2}}\dfrac{3x}{2}+{{\cos }^{2}}\left( \dfrac{\dfrac{5\pi }{3}-2x}{2} \right)=0 \\\ & \Rightarrow {{\cos }^{2}}\dfrac{3x}{2}+{{\cos }^{2}}\left( \dfrac{5\pi }{6}-x \right)=0 \\\ \end{aligned}$$ The above equation states that the sum of two squares is equal to zero. We know that the square over a term is greater than or equal to zero. Therefore, in this case, each of the above two terms must be equal to zero so that we can write $$\begin{aligned} & \Rightarrow \cos \dfrac{3x}{2}=0........\left( ii \right) \\\ & \Rightarrow \cos \left( \dfrac{5\pi }{6}-x \right)=0........\left( iii \right) \\\ \end{aligned}$$ We know that the solution to the trigonometric equation $\cos x=0$ is given by $\Rightarrow x=\left( 2n+1 \right)\dfrac{\pi }{2}$ Therefore, the solution of the equation (ii) is given by $\Rightarrow \dfrac{3x}{2}=\left( 2n+1 \right)\dfrac{\pi }{2}$ Multiplying $\dfrac{2}{3}$ both the sides, we get $$\begin{aligned} & \Rightarrow x=\left( 2n+1 \right)\dfrac{\pi }{2}\times \dfrac{2}{3} \\\ & \Rightarrow x=\left( 2n+1 \right)\dfrac{\pi }{3}........\left( iv \right) \\\ \end{aligned}$$ Similarly, the solution of the equation (iii) will be given by $\Rightarrow \dfrac{5\pi }{6}-x=\left( 2n+1 \right)\dfrac{\pi }{2}$ Multiplying the above equation by $-1$, we get $$\Rightarrow x-\dfrac{5\pi }{6}=\left( 2n+1 \right)\left( -\dfrac{\pi }{2} \right)$$ Adding $\dfrac{5\pi }{6}$ both the sides, we get $$\begin{aligned} & \Rightarrow x-\dfrac{5\pi }{6}+\dfrac{5\pi }{6}=\left( 2n+1 \right)\left( -\dfrac{\pi }{2} \right)+\dfrac{5\pi }{6} \\\ & \Rightarrow x=\left( 2n+1 \right)\left( -\dfrac{\pi }{2} \right)+\dfrac{5\pi }{6} \\\ & \Rightarrow x=\pi \left[ \left( 2n+1 \right)\left( -\dfrac{1}{2} \right)+\dfrac{5}{6} \right] \\\ & \Rightarrow x=\pi \left[ \dfrac{-3\left( 2n+1 \right)+5}{6} \right] \\\ & \Rightarrow x=\pi \left[ \dfrac{-6n-3+5}{6} \right] \\\ & \Rightarrow x=\pi \left[ \dfrac{-6n+2}{6} \right] \\\ & \Rightarrow x=\pi \left[ \dfrac{-3n+1}{3} \right] \\\ & \Rightarrow x=\left( -3n+1 \right)\dfrac{\pi }{3} \\\ \end{aligned}$$ Since n is an integer, we can replace n by negative of n to get $$\Rightarrow x=\left( 3n+1 \right)\dfrac{\pi }{3}........\left( v \right)$$ On putting different integral values of n in (iv) we get the solutions as $\Rightarrow x=\dfrac{\pi }{3},\dfrac{3\pi }{3},\dfrac{5\pi }{3},\dfrac{7\pi }{3},\dfrac{9\pi }{3},\dfrac{11\pi }{3},\dfrac{13\pi }{3}........$ Similarly, from (v) we obtain $\Rightarrow x=\dfrac{\pi }{3},\dfrac{4\pi }{3},\dfrac{7\pi }{3},\dfrac{10\pi }{3},\dfrac{13\pi }{3},.......$ From the above two solutions, we get the common solutions as $\begin{aligned} & \Rightarrow x=\dfrac{\pi }{3},\dfrac{7\pi }{3},\dfrac{13\pi }{3},....... \\\ & \Rightarrow x=\left( 6n+1 \right)\dfrac{\pi }{3},n\in z \\\ \end{aligned}$ **So, the correct answer is “Option a”.** **Note:** The given equation is not obtainable in terms of a single trigonometric term. You can try this by simplifying the LHS of the given equation using the trigonometric identities. This is because the given equation is actually the combination of two trigonometric equations. Also do not get confused between the variables k and n, since both of them denote the integer values.