The number of solutions of 2^{1+|cos x|+|cos | vedclass

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(C) Given the equation $2^{1+|\cos x|+|\cos x|^2+\ldots} = 4$. Since the exponent is an infinite geometric series with first term $a = 1$ and common r

Vedclass · Accepted Answer

$4$

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(C) Given the equation $2^{1+|\cos x|+|\cos x|^2+\ldots} = 4$. Since the exponent is an infinite geometric series with first term $a = 1$ and common ratio $r = |\cos x|$,where $|\cos x| Thus,$2^{\frac{1}{1-|\cos x|}} = 2^2$. Equating the exponents,we get $\frac{1}{1-|\cos x|} = 2$. $1 - |\cos x| = \frac{1}{2} \Rightarrow |\cos x| = \frac{1}{2}$. This implies $\cos x = \frac{1}{2}$ or $\cos x = -\frac{1}{2}$. In the interval $(-\pi, \pi)$,the solutions for $\cos x = \frac{1}{2}$ are $x = \frac{\pi}{3}, -\frac{\pi}{3}$. The solutions for $\cos x = -\frac{1}{2}$ are $x = \frac{2\pi}{3}, -\frac{2\pi}{3}$. Therefore,the total number of solutions is $4$.

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

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