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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Similar Questions

If $x=\frac{\sin^3 \theta}{\cos^2 \theta}$ and $y=\frac{\cos^3 \theta}{\sin^2 \theta}$,where $\sin \theta+\cos \theta=\frac{1}{2}$,then $x+y=$

Evaluate: $\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)$

If $\tan {3^o} + 2\tan {6^o} + 4\tan {12^o} + 8\cot {24^o} = \cot {\theta ^o}$,then:

If $A$ is in the third quadrant and $\tan A = \frac{\sqrt{7}}{3}$,then $18 - 16 \sin^2 \frac{A}{2} - 32 \sin \frac{A}{2} \sin \frac{5A}{2} = $

If $\tan A+\tan B+\cot A+\cot B=\tan A \tan B-\cot A \cot B$ and $0^{\circ} < A+B < 270^{\circ}$,then $A+B=$ (in $^{\circ}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module