Question

If sum of all the solutions of the equation $8cosx.\left( {cos\left( {\dfrac{\pi }{6} + x} \right).cos\left( {\dfrac{\pi }{6} - x} \right) - \dfrac{1}{2}} \right) = 1$in $$$$$$\left[ {0,\pi } \right]$$cis kπ, then k is equal to:

$$\;\left( 1 \right){\text{ }}\dfrac{2}{3} \\\ \left( 2 \right){\text{ }}\dfrac{{13\;}}{9} \\\ \left( 3 \right){\text{ }}\dfrac{8}{9} \\\ \left( 4 \right){\text{ }}\dfrac{{20}}{9} \\\$$

Answer 1

Solve the equation and obtain a solution and equate the sum of solution to k.
We start by sending the 2 from outside and multiply it to the bracket to get the form $2cosacosb$and then we simplify it. Then, we simplify the by taking common such that it comes in the form the trigonometry identity of$cos3x$. Then we attain a simple equation. Then we find all possible solutions which lie between the given interval. Then we find the sum of all the solutions and then we pick out solutions which lie in the given period range in the question and equate it to $k\pi$. So, we will be able to get the coefficient of k.

Complete step by step solution:
Sending the 2 inside the bracket to form $2cosacosb$
$\dfrac{{13}}{9}$
We get

$$8\cos x\left[ {\dfrac{{\cos \left( {\dfrac{\pi }{3}} \right)\cos \left( {2x} \right)}}{2} - \dfrac{1}{2}} \right] = 1 \\\ \\\$$

We take the 2 common out and cancel it with the 8 outside. Which gives us

$$4\cos x\left[ {\dfrac{1}{2} + \cos \left( {2x} \right) - 1} \right] = 1 \\\ 4\cos x\left[ {\cos (2x) - \dfrac{1}{2}} \right] = 1 \\\ 4\cos x\cos (2x) - 2\cos x = 1 \\\$$

Now again we use $2cosacosb$ trigonometric identity, such that we get
$2(cos3x + cosx) - 2cosx = 1$
The LHS equates to$cos3x$ form, So

$$\cos 3x = \dfrac{1}{2} \\\ 3x = 2n\pi \pm \dfrac{\pi }{3} \\\ x = 2n\dfrac{\pi }{3} \pm \dfrac{\pi }{9} \\\$$

Now, below are all the possible solutions for the given equations, which are

$Solutions{\text{ }}in\;[0,\pi ]\;are\;\dfrac{\pi }{9},\dfrac{{2\pi }}{3} - \dfrac{\pi }{9},2\dfrac{\pi }{3} + \dfrac{\pi }{9}$
Now, we calculate their sum and find the coefficient of k
$sum\; = \;\dfrac{\pi }{9} + \dfrac{{5\pi }}{9} + \dfrac{{7\pi }}{9} = \dfrac{{13\pi }}{9}$
K= $\dfrac{{13}}{9}$

So, the correct answer is Option 2.

Note: Here, a lot of trigonometric identities have been used. This means that good application of trigonometric identities helps solving the sums in an easier way. Also, we need to take solutions only which lie in the given period and not deviating away from the range period.