The number of elements in the set S = \ x ∈ ℝ: 2 ( x ^ 2 + x 6 ) =...

Given,

$2\cos \left(\dfrac{{x}^{2} + x}{6}\right) = {4}^{x} + {4}^{- x}$

Now we know that maximum value of $cos$ function is $1$, so $2\cos \left(\dfrac{{x}^{2} + x}{6}\right) \le 2$ and using $A . M \ge G . M$ we get ${4}^{x} + {4}^{- x} \ge 2$

So, $L . H . S . \le 2 . \text{ and } R . H . s \ge 2$

Hence $L . H . S . = 2 \text{ and } R . H . S . = 2$

Now equating both to $2$ we get,

$2\cos \left(\dfrac{{x}^{2} + x}{6}\right) = 2$ and ${4}^{x} + {4}^{- x} = 2$

Now solving $2\cos \left(\dfrac{{x}^{2} + x}{6}\right) = 2$

$⇒ \cos \left(\dfrac{{x}^{2} + x}{6}\right) = 1$

$⇒ \dfrac{{x}^{2} + x}{6} = 0$

$⇒ {x}^{2} + x = 0$

$⇒ x = 0\text{or} - 1 . . . . . . . . . . . \left(1\right)$

Now solving ${4}^{x} + {4}^{- x} = 2$ we get $x = 0 . . . . . \left(2\right)$

Now from equation \left(1\right) \text{ and } $\left(2\right)$ we can see common solution is $x = 0$

Hence, possible solution is only one.