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The number of real roots of the equation $$2\cos\left(\frac{x^2+x}6\right)=2^x+2^{-x}$$

Another question is... can we use descartes rule of sign in here or in any transcendental equation ?

Sourav
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  • Thanks you all for help.... Now please inform for academic purpose can I use Descartes rule of sign to check in any transcendental equation (if possible) or it is only for polynomial equation. – Sourav Nov 13 '13 at 10:19

3 Answers3

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The right hand side,

$$2^x + 2^{-x}$$

is an even function that has a unique minimum value of $2$ in $x = 0$. The left hand side,

$$2\cos \left( \frac{x^2+x}{6}\right)$$

has $2$ as its maximal value. So there's only one candidate, and inspection shows that it is indeed a solution.

Daniel Fischer
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Note that $2^x+2^{-x}$ has to be in the interval $[-2,2]$ to cancel the LHS. But the image of $2^x+2^{-x}$ is $[2,\infty)$, so you only have to check the case when $2^x+2^{-x}=2$, and the only solution is when $x=0$.

Integral
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Notice that

$$2\cos\left(\frac{x^2+x}6\right) \le 2$$ $$2^x+2^{-x} \ge 2$$

First is true since $\cos\left(\frac{x^2+x}6\right) \le 1$ and the second is equivalent to $\left(2^{\frac{x}{2}} - \frac{1}{2^{\frac{x}{2}}} \right)^2 \ge 0$ which is always true. So in order to have an equality the second inequality also has to be an equality and happens only when $2^{\frac{x}{2}} = \frac{1}{2^{\frac{x}{2}}} \leftrightarrow2^x=1\leftrightarrow x = 0$.

sve
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  • Notice that $2\ge 2$ and $2\sin x \le 2$ for all $x$. Yet $2\sin x =2$ has quite a few more solutions that the OP's equation. – Fly by Night Nov 12 '13 at 21:13
  • @FlybyNight that was just a hint... it was supposed to make him explorer the corner cases. – sve Nov 12 '13 at 21:21