The number of solutions, of $2^{1+|\cos x|+|\cos x|^2+\ldots \ldots \ldots \cdots}=4$ in $(-\pi, \pi)$, is

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Aicharya · Accepted Answer

$\begin{aligned} & 2^{1+|\cos x|+|\cos x|^2+\ldots}=4 \ & \Rightarrow 2^{\frac{1}{1-|\cos x|}}=2^2\end{aligned}$ $\begin{aligned} & \Rightarrow \frac{1}{1-|\cos x|}=2 \ & \Rightarrow 1-|\cos x|=\frac{1}{2} \ & \Rightarrow|\cos x|=\frac{1}{2} \ & \Rightarrow \cos x= \pm \frac{1}{2} \ & \Rightarrow x=\frac{-2 \pi}{3}, \frac{-\pi}{3}, \frac{\pi}{3}, \frac{2 \pi}{3} \quad \ldots[\because x \in(-\pi, \pi)] \end{aligned}$ $	herefore \quad$ Number of solutions $=4$

The number of solutions, of $2^{1+|\cos x|+|\cos x|^2+\ldots \ldots \ldots \cdots}=4$ in $(-\pi, \pi)$, is

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