The values of x in (-pi, pi) which satisfy th | vedclass

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(A) The given equation is $8^{1+\cos ^2 x+\cos ^4 x+\ldots}=4^3$. Since the exponent is an infinite geometric series with first term $a=1$ and common

Vedclass · Accepted Answer

$\pm \frac{\pi}{4}, \pm \frac{3 \pi}{4}$

Vedclass · Answer

(A) The given equation is $8^{1+\cos ^2 x+\cos ^4 x+\ldots}=4^3$. Since the exponent is an infinite geometric series with first term $a=1$ and common ratio $r=\cos^2 x$,where $|\cos^2 x| Substituting this into the equation: $8^{\frac{1}{\sin^2 x}} = 4^3$. Expressing both sides with base $2$: $(2^3)^{\frac{1}{\sin^2 x}} = (2^2)^3$. $2^{\frac{3}{\sin^2 x}} = 2^6$. Equating the exponents: $\frac{3}{\sin^2 x} = 6$. $\sin^2 x = \frac{3}{6} = \frac{1}{2}$. Taking the square root: $\sin x = \pm \frac{1}{\sqrt{2}}$. For $x \in (-\pi, \pi)$,the solutions are $x = \pm \frac{\pi}{4}, \pm \frac{3\pi}{4}$.

The values of $x$ in $(-\pi, \pi)$ which satisfy the equation $8^{1+\cos ^2 x+\cos ^4 x+\ldots}=4^3$ are

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