If sum of all the solutions of the equation [8 x.( ( 6 + x ). ( 6 - x ) - 1 2 ) = 1] in [0…

Question

If sum of all the solutions of the equation

$$8\cos x.\left( {\cos \left( {{\pi  \over 6} + x} \right).\cos \left( {{\pi  \over 6} - x} \right) - {1 \over 2}} \right) = 1$$

in [0, $$\pi $$] is k$$\pi $$, then k is equal to

Accepted Answer

As we know,

$$\cos \left( {x + y} \right)\cos \left( {x - y} \right) = {\cos ^2}x - {\sin ^2}y$$

$$\therefore$$ $${\mathop{\rm cos}\nolimits} \left( {{\pi  \over 6} + x} \right)cos\left( {{\pi  \over 6} - x} \right) = {\cos ^2}\left( {{\pi  \over 6}} \right) - {\sin ^2}x$$

Given, $$8\cos x\left( {\cos \left( {{\pi  \over 6} + x} \right)\cos \left( {{\pi  \over 6} - x} \right) - {1 \over 2}} \right) = 1$$

$$ \Rightarrow 8\cos x\left( {{{\cos }^2}{\pi  \over 6} - {{\sin }^2}x - {1 \over 2}} \right) = 1$$

$$ \Rightarrow 8\cos x\left( {{3 \over 4} - {1 \over 2} - {{\sin }^2}x} \right) = 1$$

$$ \Rightarrow 8\cos x\left( {{3 \over 4} - {1 \over 2} - 1 + {{\cos }^2}x} \right) = 1$$

$$ \Rightarrow 8\cos x\left( {{{\cos }^2}x - {3 \over 4}} \right) = 1$$

$$ \Rightarrow 2\left( {4{{\cos }^3}x - 3\cos x} \right) = 1$$

(Taking $$4\cos x$$ inside the bracket).

We know,

$$4{\cos ^3}x - 3\cos x = \cos 3x$$

$$ \Rightarrow 2\cos 3x = 1$$

$$ \Rightarrow \cos 3x = {1 \over 2}$$

$$\therefore$$ $$\,\,\,$$ $$3x = 2n\pi  \pm {\pi  \over 3}$$

So, $$x = {{2n\pi } \over 3} \pm {\pi  \over 9}$$

At $n=0,$ $$x =  + {\pi  \over 9}$$ as $$x \in \left[ {0,\pi } \right]$$

At $n=1,$ $$x = {{2\pi } \over 3} + {\pi  \over 9}$$

$$\therefore$$ $$\,\,\,$$ $$x = {{2\pi } \over 3} + {\pi  \over 9}$$

and $$x = {{2\pi } \over 3} - {\pi  \over 9}$$

At $n = 2,$ $$x = {{4\pi } \over 3} \pm {\pi  \over 9}$$

Which does not belongs to $$\left[ {0,\pi } \right]$$

So, possible values of $x$ is $$ = {\pi  \over 9},{{2\pi } \over 3} + {\pi  \over 9},{{2\pi } \over 3} - {\pi  \over 3}$$

Sum of the solutions $$ = {\pi  \over 9} + {{2\pi } \over 3} + {\pi  \over 9} + {{2\pi } \over 3} - {\pi  \over 9}$$ v

$$ = {{4\pi } \over 3} + {\pi  \over 9}$$ $$ = {{13\,\pi } \over 9}$$

According to the question,

$$k\pi  = {{13\pi } \over 9}$$

$$\therefore\,\,\,$$ $$k = {{13} \over 9}$$

Solutions · 1