QJEEM18Ma89cc: JEE Main 2018 Mathematics - Class 11 JEEM18Ma89cc Solution: If sum of all the solutions of the equation in [0, ] is k ,…

Question

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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}

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