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Find the number of real roots of

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

1) 0

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2) 1

3) 2

4) None of these

My guess is to approach it in graphical way. But equation seems little difficult.

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HINT:

For real $x,\dfrac{2^x+2^{-x}}2\ge\sqrt{2^x\cdot 2^{-x}}=1$

and for real $y,\cos y\le1$

Rearranging we get $$(2^x)^2-2\cos\dfrac{x^2+x}6(2^x)+1=0\ \ \ \ (0)$$ which is a Quadratic equation in $2^x$

For real $2^x,$ the discriminant must be $\ge0$

i.e., $$\left(2\cos\dfrac{x^2+x}6\right)^2-4=-4\sin^2\dfrac{x^2+x}6\ge0\ \ \ \ (1)$$

As $\sin\dfrac{x^2+x}6$ is real, $$\sin^2\dfrac{x^2+x}6\ge0\iff-4\sin^2\dfrac{x^2+x}6\le0\ \ \ \ (2)$$

By $(1),(2);$ $$\sin^2\dfrac{x^2+x}6=0\implies\cos\dfrac{x^2+x}6=\pm1$$

What happens to $(0)$ if $(i):\cos\dfrac{x^2+x}6=1$

Or if $(ii):\cos\dfrac{x^2+x}6=-1$

Remember for real $x, 2^x>0$

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