Question

The value of $x$ which satisfies ${8^{1 + \cos x + {{\cos }^2}x + ....}} = 64$ in $\left[ { - \pi ,\pi } \right]$ is
A.$\pm \dfrac{\pi }{2}, \pm \dfrac{\pi }{3}$
B.$\pm \dfrac{\pi }{3}$
C.$\pm \dfrac{\pi }{2}, \pm \dfrac{\pi }{6}$
D.$\pm \dfrac{\pi }{6}, \pm \dfrac{\pi }{3}$

Answer 1

First of all we will write 64 as ${\left( 8 \right)^2}$ and then we will compare the equation and simplify it. Next, we will see an infinite G.P. Apply the sum of infinite terms of an G.P. to further simplify it. Next, solve the formed equation to find the value of $x$.

Complete step-by-step answer:
We have been given that ${8^{1 + \cos x + {{\cos }^2}x + ....}} = 64$
We can write 64 as ${\left( 8 \right)^2}$
Then, we will have ${8^{1 + \cos x + {{\cos }^2}x + ....}} = {8^2}$
On comparing the equation, we will get ,
$1 + \cos x + {\cos ^2}x + .... = 2$
Here, we can see the series on the left side is an infinite G.P. where the first term is 1 and the common ratio is $\cos x$
We know that the sum of infinite terms of a G.P. is given as $\dfrac{a}{{1 - r}}$, where $a$ is the first term, $r$ is the common ratio and $\left| r \right| < 1$
Now, we have to consider only when $\left| {\cos x} \right| < 1$, that is, $\left| {\cos x} \right| \ne 1$
Then, $1 + \cos x + {\cos ^2}x + .... = 2$ can be simplified as,
$\dfrac{1}{{1 - \cos x}} = 2 \\\ \Rightarrow 1 = 2 - 2\cos x \\\ \Rightarrow 2\cos x = 1 \\\ \Rightarrow \cos x = \dfrac{1}{2} \\\$
Now, we know that when $\cos x = \dfrac{1}{2} \Rightarrow x = - \dfrac{\pi }{3},\dfrac{\pi }{3}$
Hence, the value of $x$ is $\pm \dfrac{\pi }{3}$
Thus, option B is correct.

Note: A geometric progression (G.P) is a sequence in which we get terms by multiplying a common ratio to the preceding term. We must note that whenever we use the formula of infinite terms of G.P $\dfrac{a}{{1 - r}}$, we have to consider $\left| r \right| < 1$. Also, if we want to find the sum of finite terms of G.P., we use ${S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$, where $a$ is the first term, $r$ is the common ratio and $n$ is the number of terms.