How to find the number of real roots of the given equation? 
The number of real roots of the equation $$2 \cos \left( \frac{x^2+x}{6} \right)=2^x+2^{-x}$$ is
  (A) $0$, (B) $1$, (C) $2$, (D) infinitely many.

Trial:  $$\begin{align} 2 \cos \left( \frac{x^2+x}{6} \right)&=2^x+2^{-x} \\ \implies \frac{x^2+x}{6}&=\cos ^{-1} \left( \frac{2^x+2^{-x}}{2} \right)\end{align}$$ Then I can't proceed.
 A: There's a clever approach to this problem. In particular, $$(w-1)^2 \geq 0 \implies w^2 + 1 \geq 2w \implies w + \frac{1}{w} \geq 2$$
and this works for any $w$. Using this, you can show that one side is always at least $2$. Meanwhile, the other is at most $2$.
A: $$\text{So, }(2^x)^2- 2^x\cdot2\cos\left(\frac{x^2+x}6\right)+1=0$$ which is a quadratic equation in $2^x$ 
As $x$ is real, so would be $2^x$ 
so,the discriminant must be $\ge 0$ 
i.e., $$\{2 \cos\left(\frac{x^2+x}6\right)\} ^2-4\cdot1\cdot1\ge0$$
$$\implies4\left(\cos^2\left(\frac{x^2+x}6\right)-1\right)\ge0$$
$$\implies  \sin^2\left(\frac{x^2+x}6\right)\le 0$$
$$\implies  \sin^2\left(\frac{x^2+x}6\right)= 0\text{ as the square of a real number }\ge 0 $$
So, the discriminant must be $0,\implies \cos\left(\frac{x^2+x}6\right)=\pm1$
$\implies 2^x=\frac{2\cos\left(\frac{x^2+x}6\right)\pm\sqrt{2 \cos\left(\frac{x^2+x}6\right)\}^2-4\cdot1\cdot1}}2=\pm1$
As $x$ is real, $2^x>0$
$\implies 2^x=1\implies x=0$ which also satisfies $ \sin^2\left(\frac{x^2+x}6\right)= 0$
A: If you consider $f(x)=2^x+2^{-x}$ and notice how it is symmetrical around $x=0$ and has its minimum there as for example $f(1)=2.5$ and $f(2)=4.25$, which the left hand side has a maximum value of 2, thus there is only one possible root though I'll leave that for you to see.
A: Notice that $2^0+2^0=2$ and that $2^x+2^{-x}> 2$ for $x\ne 0$.  Then consider that $2\cos(0)=2$ and that $2\cos\left(\frac{x^2+x}{6}\right)\le2$ for all $x$.  What can we conclude?
A: *

*$RHS=2^x+2^{-x}=2^x+\frac{1}{2^x}\geq 2$ by $AM\geq GM$, since $2^x>0$ for all $x\in\mathbb{R}$

*$LHS\leq 2$ for all $x\in\mathbb{R}$. This is obvious.

*One can verify $x=0$  is a solution. 

*$f(x)=2^x+2^{-x}$ implies $f'(x)>0$ for $x>0$. 
These should suffice to conclude the unique solution at $x=0$. 
