Given, $2\cos \left( {{{{x^2} + x} \over 6}} \right) = {4^x} + {4^{ - x}}$ We know, A.M $\ge$ G.M $\therefore$ for 4 x and 4 $-$x ${{{4^x} + {4^{ - x}}} \over 2} \ge \sqrt {{4^x}\,.\,{4^{ - x}}}$ $\Rightarrow {{{4^x} + {4^{ - x}}} \over 2} \ge 1$ $\Rightarrow {4^x} + {4^{ - x}} \ge 2$ ...... (1) And we know, $- 1 \le \cos x \le 1$ $\therefore$ $- 1 \le \cos \left( {{{{x^2} + x} \over 6}} \right) \le 1$ $- 2 \le 2\cos \left( {{{{x^2} + x} \over 6}} \right) \le 2$ ...... (2) (1) and (2) both satisfies only when both equal to 2. $\therefore$ $2\cos \left( {{{{x^2} + x} \over 6}} \right) = 2$ $\Rightarrow \cos \left( {{{{x^2} + x} \over 6}} \right) = 1$ $\Rightarrow \cos \left( {{{{x^2} + x} \over 6}} \right) = \cos 0$ $\Rightarrow {{{x^2} + x} \over 6} = 0$ $\Rightarrow {x^2} + x = 0$ $\Rightarrow x(x + 1) = 0$ $\Rightarrow x = 0, - 1$ When $x = 0$, L.H.S $= 2\cos \left( {{{{x^2} + x} \over 6}} \right)$ $= 2\cos \left( {{0 \over 6}} \right)$ $= 2\cos 0$ $= 2\,.\,1$ $= 2$ R.H.S $= {4^x} + {4^{ - x}}$ $= {4^0} + {4^0}$ $= 2$ $\therefore$ $x = 0$ is accepted. Now, when $x = - 1$, L.H.S $= 2\cos \left( {{{{x^2} + x} \over 6}} \right)$ $= 2\cos \left( {{{1 - 1} \over 6}} \right)$ $= 2\cos 0 = 2$ R.H.S $= {4^x} + {4^{ - x}}$ $= {4^{ - 1}} + {4^1}$ $= {1 \over 4} + 4$ $= {{17} \over 4}$ $\therefore$ L.H.S $\ne$ R.H.S $\therefore$ $x = - 1$ is not a solution. $\therefore$ Only one solution possible which is $x = 0$.