Question

If the sum of all the solutions of the equation $8\cos x\cdot \left( \cos \left( \dfrac{\pi }{6}+x \right)\cdot \cos \left( \dfrac{\pi }{6}-x \right)-\dfrac{1}{2} \right)=1$ in $\left[ 0,\pi \right]$ is $k\pi$ , then $k$ is equal to.
(a) $\dfrac{8}{9}$
(b) $\dfrac{20}{9}$
(c) $\dfrac{2}{3}$
(d) $\dfrac{13}{9}$

Answer 1

Hint: For solving this question we will use some trigonometric formulae and then find the value of $x$ which will satisfy the given equation. After finding the solutions we will just add them and find the value of $k$.

Complete step-by-step solution -
Given:
$8\cos x\cdot \left( \cos \left( \dfrac{\pi }{6}+x \right)\cdot \cos \left( \dfrac{\pi }{6}-x \right)-\dfrac{1}{2} \right)=1$ and $x\in \left[ 0,\pi \right]$
It is given that $x$ can take values from $0$ to $\pi$ including $0$ & $\pi$ .
We will use the following formulae in solving this question:
$\begin{aligned} & \cos \left( A+B \right)=\cos A\cdot \cos B-\sin A\cdot \sin B..........\left( 1 \right) \\\ & \cos \left( A-B \right)=\cos A\cdot \cos B+\sin A\cdot \sin B..........\left( 2 \right) \\\ & \cos \left( 3\theta \right)=4{{\cos }^{3}}\theta -3\cos \theta .............................\left( 3 \right) \\\ & \cos \dfrac{\pi }{6}=\dfrac{\sqrt{3}}{2}.....................................................\left( 4 \right) \\\ & \sin \dfrac{\pi }{6}=\dfrac{1}{2}........................................................\left( 5 \right) \\\ & \left( a+b \right)\times \left( a-b \right)={{a}^{2}}-{{b}^{2}}................................\left( 6 \right) \\\ & {{\cos }^{2}}x+{{\sin }^{2}}x=1...........................................\left( 7 \right) \\\ \end{aligned}$

We have,
$8\cos x\cdot \left( \cos \left( \dfrac{\pi }{6}+x \right)\cdot \cos \left( \dfrac{\pi }{6}-x \right)-\dfrac{1}{2} \right)=1$
Using (1) in the above equation,
$8\cos x\cdot \left( \left( \cos \dfrac{\pi }{6}\cdot \cos x-\sin \dfrac{\pi }{6}\cdot \sin x \right)\times \left( \cos \dfrac{\pi }{6}\cdot \cos x+\sin \dfrac{\pi }{6}\cdot \sin x \right)-\dfrac{1}{2} \right)=1$
Now, using (4) and (5) in the above equation,

& 8\cos x\cdot \left( \left( \dfrac{\sqrt{3}}{2}\cdot \cos x-\dfrac{1}{2}\cdot \sin x \right)\times \left( \dfrac{\sqrt{3}}{2}\cdot \cos x+\dfrac{1}{2}\cdot \sin x \right)-\dfrac{1}{2} \right)=1 \\\ & \Rightarrow 8\cos x\cdot \left( \left( \dfrac{\sqrt{3}\cos x}{2}-\dfrac{\sin x}{2} \right)\times \left( \dfrac{\sqrt{3}\cos x}{2}+\dfrac{\sin x}{2} \right)-\dfrac{1}{2} \right)=1 \\\ \end{aligned}$$ Now, using (6) in the above equation, $$\begin{aligned} & 8\cos x\cdot \left( {{\left( \dfrac{\sqrt{3}\cos x}{2} \right)}^{2}}-{{\left( \dfrac{\sin x}{2} \right)}^{2}}-\dfrac{1}{2} \right)=1 \\\ & \Rightarrow 8\cos x\cdot \left( \dfrac{3{{\cos }^{2}}x}{4}-\dfrac{{{\sin }^{2}}x}{4}-\dfrac{1}{2} \right)=1 \\\ & \Rightarrow 8\cos x\cdot \left( \dfrac{3{{\cos }^{2}}x}{4}+\dfrac{{{\cos }^{2}}x}{4}-\dfrac{{{\cos }^{2}}x}{4}-\dfrac{{{\sin }^{2}}x}{4}-\dfrac{1}{2} \right)=1 \\\ & \Rightarrow 8\cos x\cdot \left( {{\cos }^{2}}x-\left( \dfrac{{{\cos }^{2}}x+{{\sin }^{2}}x}{4} \right)-\dfrac{1}{2} \right)=1 \\\ \end{aligned}$$ Now, using (7) in the above equation, $$\begin{aligned} & 8\cos x\cdot \left( {{\cos }^{2}}x-\dfrac{1}{4}-\dfrac{1}{2} \right)=1 \\\ & \Rightarrow 8\cos x\cdot \left( {{\cos }^{2}}x-\dfrac{3}{4} \right)=1 \\\ & \Rightarrow 2\left( 4\cos x\times {{\cos }^{2}}x-4\cos x\times \dfrac{3}{4} \right)=1 \\\ & \Rightarrow 2\left( 4{{\cos }^{3}}x-3\cos x \right)=1 \\\ \end{aligned}$$ Now, using (3) in the above equation, $\begin{aligned} & 2\times \cos \left( 3x \right)=1 \\\ & \Rightarrow \cos \left( 3x \right)=\dfrac{1}{2}...............\left( 8 \right) \\\ \end{aligned}$ Now, before we proceed we should know one important result which we will use here. If $\cos x=\cos y$ , then general solution for $x$ in terms of y can be written as, $x=2n\pi \pm y............\left( 9 \right)$ , where $n$ is any integer. Before we find $x$ , we should remember that $x$ can take values from $0$ to $\pi $ including $0$ & $\pi $ . From (8) we have: $\cos 3x=\dfrac{1}{2}$ We know that, $\cos \dfrac{\pi }{3}=\dfrac{1}{2}$ . Then, $\cos 3x=\cos \dfrac{\pi }{3}$ Now, using (9) in the above equation, $\begin{aligned} & \cos 3x=\cos \dfrac{\pi }{3} \\\ & \Rightarrow 3x=2n\pi \pm \dfrac{\pi }{3} \\\ & \Rightarrow x=\dfrac{2n\pi }{3}\pm \dfrac{\pi }{9} \\\ \end{aligned}$ First consider, $x=\dfrac{2n\pi }{3}+\dfrac{\pi }{9}$ , Put, $n=0$ . Then, $\begin{aligned} & x=0+\dfrac{\pi }{9} \\\ & \Rightarrow x=\dfrac{\pi }{9}...........\left( 10 \right) \\\ \end{aligned}$ Put, $n=1$ . Then, $\begin{aligned} & x=\dfrac{2\pi }{3}+\dfrac{\pi }{9} \\\ & \Rightarrow x=\dfrac{7\pi }{9}..........\left( 11 \right) \\\ \end{aligned}$ Put, $n=2$ . Then, $\begin{aligned} & x=\dfrac{4\pi }{3}+\dfrac{\pi }{9} \\\ & \Rightarrow x=\dfrac{13\pi }{9}\left( >\pi \right) \\\ \end{aligned}$ Now consider, $x=\dfrac{2n\pi }{3}-\dfrac{\pi }{9}$ Put, $n=1$ . Then, $\begin{aligned} & x=\dfrac{2\pi }{3}-\dfrac{\pi }{9} \\\ & \Rightarrow x=\dfrac{5\pi }{9}.............\left( 12 \right) \\\ \end{aligned}$ Put, $n=2$ . Then, $\begin{aligned} & x=\dfrac{4\pi }{3}-\dfrac{\pi }{9} \\\ & \Rightarrow x=\dfrac{11\pi }{9}\left( >\pi \right) \\\ \end{aligned}$ In above calculations we neglected the values which are greater than $\pi $ or lesser than 0. From (10), (11) and (12) we have: $\begin{aligned} & x=\dfrac{\pi }{9} \\\ & x=\dfrac{7\pi }{9} \\\ & x=\dfrac{5\pi }{9} \\\ \end{aligned}$ Then, sum of all the solutions $=\dfrac{\pi }{9}+\dfrac{7\pi }{9}+\dfrac{5\pi }{9}=\dfrac{13\pi }{9}=\dfrac{13}{9}\times \pi $ . It is given that, sum of all the solutions $=k\pi $ . Then, $k=\dfrac{13}{9}$ . Thus, (d) is the correct option. Note: Here, we should use each formula which is mentioned in the solution. And for finding the value of $x$ , we should remember that $x$ can take values from $0$ to $\pi $ including $0$ & $\pi $ . Then, calculate the value of $k$ and match the correct option.