Question

$\tt \red{The \: number \: of \: real} \\ \tt \color{blue}{ solution \: of \: the \: equation} \\ \tiny\tt \pink{\sin^{ - 1} \bigg( \sum \limits_{i = 1}^{ \infty } {x}^{i + 1} - x \sum \limits_{i = 1}^{ \infty } ( \frac{x}{2} {)}^{i} \bigg) = \frac{\pi}{2} - {cos}^{ - 1} \bigg( \sum \limits _{ i = 1 }^{\infty} ( \frac{ - x}{2} ) ^{i} - \sum\limits_{ i = 1 }^{\infty} ( - x {)}^{i} \bigg)} \\ \sf \color{green}{ lying \: in \: the \: interval \: \bigg( - \frac{1}{2}, \frac{1}{2} \bigg) \: is}$​

Answer 1

$\large\underline{\sf{Solution-}}$

Given inverse Trigonometric equation is

$\rm :\longmapsto\:\sf{\sin^{ - 1} \bigg( \sum \limits_{i = 1}^{ \infty } {x}^{i + 1} - x \sum \limits_{i = 1}^{ \infty } ( \frac{x}{2} {)}^{i} \bigg) = \dfrac{\pi}{2} - {cos}^{ - 1} \bigg( \sum \limits _{ i = 1 }^{\infty} ( \frac{ - x}{2} ) ^{i} - \sum\limits_{ i = 1 }^{\infty} ( - x {)}^{i} \bigg)}$

can be further rewritten as

$\rm :\longmapsto\:\sf{\sin^{ - 1} \bigg( \sum \limits_{i = 1}^{ \infty } {x}^{i + 1} - x \sum \limits_{i = 1}^{ \infty } ( \frac{x}{2} {)}^{i} \bigg) + {cos}^{ - 1} \bigg( \sum \limits _{ i = 1 }^{\infty} ( \frac{ - x}{2} ) ^{i} - \sum\limits_{ i = 1 }^{\infty} ( - x {)}^{i} \bigg)} = \dfrac{\pi}{2}$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ {sin}^{ - 1}x + {cos}^{ - 1}x = \dfrac{\pi}{2}}}}$

So, using this identity, we get

$\rm :\longmapsto\:\sf{ \sum \limits_{i = 1}^{ \infty } {x}^{i + 1} - x \sum \limits_{i = 1}^{ \infty } ( \frac{x}{2} {)}^{i}= \sum \limits _{ i = 1 }^{\infty} ( \frac{ - x}{2} ) ^{i} - \sum\limits_{ i = 1 }^{\infty} ( - x {)}^{i}}$

Now, Consider

$\rm :\longmapsto\:\sf{ \sum \limits_{i = 1}^{ \infty } {x}^{i + 1}}$

$\rm \:  =  \: {x}^{2} + {x}^{3} + {x}^{4} + - - - \infty$

Its an infinite GP series, so using sum of infinite GP series, we get

$\rm \:  =  \: \dfrac{ {x}^{2} }{1 - x}$

Now, Consider

$\rm :\longmapsto\:x \sum \limits_{i = 1}^{ \infty } \bigg( \dfrac{x}{2}\bigg)^{i}$

$\rm \:  =  \: x\bigg[\dfrac{x}{2} + \dfrac{ {x}^{2} }{4} + \dfrac{ {x}^{3} }{8 } + - - - \infty \bigg]$

Its an infinite GP series, So using sum of infinite GP series, we get

$\rm \:  =  \: x \times \dfrac{\dfrac{x}{2} }{1 - \dfrac{x}{2} }$

$\rm \:  =  \: \dfrac{ {x}^{2} }{2 - x}$

Now, Consider

$\rm :\longmapsto\: \sum \limits_{i = 1}^{ \infty } \bigg( - \dfrac{x}{2}\bigg)^{i}$

$\rm \:  =  \: - \dfrac{x}{2} + \dfrac{ {x}^{2} }{4} - \dfrac{ {x}^{3} }{8 } + - - - \infty$

$\rm \:  =  \: \dfrac{ - \dfrac{x}{2} }{1 + \dfrac{x}{2} }$

$\rm \:  =  \: \dfrac{ - x}{2 + x}$

Now, Consider

$\rm :\longmapsto\:\sf{ \sum \limits_{i = 1}^{ \infty } {( - x)}^{i}}$

$\rm \:  =  \: - x + {x}^{2} - {x}^{3} + {x}^{4} + - - - \infty$

Its an infinite GP series, so using sum of infinite GP series, we get

$\rm \:  =  \: \dfrac{ - x}{1 + x}$

So, on substituting these values in above expression, we get

$\rm :\longmapsto\:\dfrac{ {x}^{2} }{1 - x} - \dfrac{ {x}^{2} }{2 - x} = \dfrac{x}{1 + x} - \dfrac{x}{2 + x}$

$\rm :\longmapsto\:\dfrac{ {x}^{2} }{1 - x} - \dfrac{ {x}^{2} }{2 - x} - \dfrac{x}{1 + x} + \dfrac{x}{2 + x} = 0$

$\rm :\longmapsto\:x\bigg[\dfrac{ {x}}{1 - x} - \dfrac{ {x}}{2 - x} - \dfrac{1}{1 + x} + \dfrac{1}{2 + x}\bigg] = 0$

$\bf\implies \:x = 0$

and

$\rm :\longmapsto\:\dfrac{ {x}}{1 - x} - \dfrac{ {x}}{2 - x} - \dfrac{1}{1 + x} + \dfrac{1}{2 + x} = 0$

$\rm :\longmapsto\:\dfrac{ {x}}{1 - x} - \dfrac{1}{1 + x} = \dfrac{x}{2 - x} - \dfrac{1}{2 + x}$

$\rm :\longmapsto\:\dfrac{ {x(1 + x) - 1 + x}}{1 - {x}^{2} } = \dfrac{x(2 + x) - 2 + x}{4 - {x}^{2} }$

$\rm :\longmapsto\:\dfrac{ {x + {x}^{2} - 1 + x}}{1 - {x}^{2} } = \dfrac{2x + {x}^{2} - 2 + x}{4 - {x}^{2} }$

$\rm :\longmapsto\:\dfrac{ {{x}^{2} - 1 + 2x}}{1 - {x}^{2} } = \dfrac{{x}^{2} + 3x - 2}{4 - {x}^{2} }$

$\rm :\longmapsto\:( {x}^{2} + 2x - 1)(4 - {x}^{2}) = (1 - {x}^{2})( {x}^{2} + 3x - 2)$

$\rm :\longmapsto\: {4x}^{2} + 8x - 4 - {x}^{4} - {2x}^{3} + {x}^{2} = {x}^{2} + 3x - 2 - {x}^{4} - {3x}^{3} + {2x}^{2}$

$\rm :\longmapsto\: {x}^{3} + {2x}^{2} + 5x - 2 = 0$

Now, its a cubic equation, to find the nature of root of it, we use the concept of increasing and decreasing.

Let assume that

$\rm :\longmapsto\: p(x) = {x}^{3} + {2x}^{2} + 5x - 2$

So,

$\rm :\longmapsto\: p'(x) = 3{x}^{2} + 4x + 5$

Now, Its a quadratic equation, whose a > 0 and Discriminant, D = 16 - 60 = - 44 < 0

$\bf\implies \:p'(x) > 0$

$\bf\implies \:p(x) \: is \: increasing \: function$

$\bf\implies \:p(x) \: intersects \:x \: - \: axis \: only \: once$

As, we know that every cubic equation has atleast one real root.

Now,

$\rm :\longmapsto\:p(0) = 0 + 0 + 0 - 2 = - 2$

and

$\rm :\longmapsto\:p(0.5) = 0.625 + 0.5 + 2.5 - 2 = 1.125$

$\bf\implies \:p(x) \: has \: one \: root \: between \: \bigg(0, \: \dfrac{1}{2} \bigg)$

$\bf\implies \:Number \: of \: real \: roots \: are \: 2$

Answer 2

Step-by-step explanation:

Hey mate

Here is the answer

EXPLANATION.

${\displaystyle {\begin{aligned}\tan x&{}=\sum _{n=0}^{\infty }{\frac {U_{2n+1}}{(2n+1)!}}x^{2n+1}\\[8mu]&{}=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}2^{2n}\left(2^{2n}-1\right)B_{2n}}{(2n)!}}x^{2n-1}\\[5mu]&{}=x+{\frac {1}{3}}x^{3}+{\frac {2}{15}}x^{5}+{\frac {17}{315}}x^{7}+\cdots ,\qquad {\text{for }}|x|<{\frac {\pi }{2}}.\end{aligned}}}{\displaystyle {\begin{aligned}\tan x&{}=\sum _{n=0}^{\infty }{\frac {U_{2n+1}}{(2n+1)!}}x^{2n+1}\\[8mu]&{}=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}2^{2n}\left(2^{2n}-1\right)B_{2n}}{(2n)!}}x^{2n-1}\\[5mu]&{}=x+{\frac {1}{3}}x^{3}+{\frac {2}{15}}x^{5}+{\frac {17}{315}}x^{7}+\cdots ,\qquad {\text{for }}|x|<{\frac {\pi }{2}}.\end{aligned}}}$

${\displaystyle {\begin{aligned}\sec x&=\sum _{n=0}^{\infty }{\frac {U_{2n}}{(2n)!}}x^{2n}=\sum _{n=0}^{\infty }{\frac {(-1)^{n}E_{2n}}{(2n)!}}x^{2n}\\[5mu]&=1+{\frac {1}{2}}x^{2}+{\frac {5}{24}}x^{4}+{\frac {61}{720}}x^{6}+\cdots ,\qquad {\text{for }}|x|<{\frac {\pi }{2}}.\end{aligned}}}$

${\displaystyle {\begin{aligned}\cot x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}2^{2n}B_{2n}}{(2n)!}}x^{2n-1}\\[5mu]&=x^{-1}-{\frac {1}{3}}x-{\frac {1}{45}}x^{3}-{\frac {2}{945}}x^{5}-\cdots ,\qquad {\text{for }}0<|x|<\pi .\end{aligned}}}$

Parity:

The cosine and the secant are even functions; the other trigonometric functions are odd functions. That is:

${\displaystyle {\begin{aligned}\sin(-x)&=-\sin x\\\cos(-x)&=\cos x\\\tan(-x)&=-\tan x\\\cot(-x)&=-\cot x\\\csc(-x)&=-\csc x\\\sec(-x)&=\sec x.\end{aligned}}}$

Periods :

All trigonometric functions are periodic functions of period 2π. This is the smallest period, except for the tangent and the cotangent, which have π as smallest period. This means that, for every integer k, one has

${\displaystyle {\begin{aligned}\sin(x+2k\pi )&=\sin x\\\cos(x+2k\pi )&=\cos x\\\tan(x+k\pi )&=\tan x\\\cot(x+k\pi )&=\cot x\\\csc(x+2k\pi )&=\csc x\\\sec(x+2k\pi )&=\sec x.\end{aligned}}}$

Table of contents::

${\displaystyle {\begin{array}{|c|c|c|}\hline f(x)&f'(x)&\int f(x)\,dx\\\hline \sin x&\cos x&-\cos x+C\\\cos x&-\sin x&\sin x+C\\\tan x&\sec ^{2}x=1+\tan ^{2}x&-\ln |\cos x|+C\\\csc x&-\csc x\cot x&-\ln |\csc x+\cot x|+C\\\sec x&\sec x\tan x&\ln |\sec x+\tan x|+C\\\cot x&-\csc ^{2}x=-1-\cot ^{2}x&\ln |\sin x|+C\\\hline \end{array}}}$

Note:

Please scroll to see full answer

Regards

Itzintellectual