Question

If $\alpha$ is the value of $xϵ[0,2\pi ]$ which is a solution of the equation

2 $co{s}^2$ $\frac{x^2+x}{6}$ = $2^x$ + $2^{-x}$, then find the value of $\frac{2\alpha }{3\pi }$ .

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Answer 1

Maximum value of 2 $co{s}^2$$\frac{x^2+x}{6}$ is 2. Minimum value of $2^x$ + $2^{-x}$ is 2. We can prove this by the relation A.M ≥ G.M

$\Rightarrow$ $\frac{\left(2^x\right)+\left(2^{-x}\right)}{2}$ ≥ $\sqrt{2^x{2}^{-x}}$

$\Rightarrow$ $2^x$ + $2^{-x}$ ≥ 2

2 cos$\frac{x^2+x}{6}$ ≤ 2 and 2 $co{s}^2$$\frac{x^2+x}{6}$ = $2^x$ + $2^{-x}$ ≥ 2

$\Rightarrow$ 2 ≤ 2 cos$\frac{x^2+x}{6}$ ≤ 2

$\Rightarrow$ 2 cos$\frac{x^2+x}{6}$ = 2

$\Rightarrow$ cos$\frac{x^2+x}{6}$ = 1 ....................(1)

$\Rightarrow$ $2^x$ + $2^{-x}$ = 2 cos$\frac{x^2+x}{6}$ = 2

$\Rightarrow$ $2^x$ + $2^{-x}$ = 2

$\Rightarrow$ x = 0 [we can find this by substituting $2^x$ = 0 and forming a quadratic ]

x = 0 satisfies (1) also