The sum of the values of $x$ satisfying the equation $\sin ^{-1}\left(\frac{3 x}{5}\right)+\sin…
The sum of the values of $x$ satisfying the equation $\sin ^{-1}\left(\frac{3 x}{5}\right)+\sin ^{-1}\left(\frac{4 x}{5}\right)=\sin ^{-1}(x)$, is
- 0
- $\frac{1}{2}$
- $\frac{1}{3}$
- $\frac{1}{4}$
Solution
Given, equation
$\begin{aligned}
& \sin ^{-1}\left(\frac{3 x}{5}\right)+\sin ^{-1} \frac{4 x}{5}=\sin ^{-1} x, x \in[-1,1] \\
& \therefore \sin ^{-1}\left(\frac{3 x}{5} \sqrt{1-\frac{16 x^2}{25}}+\frac{4 x}{5} \sqrt{1-\frac{9 x^2}{25}}\right)=\sin ^{-1} x \\
& \Rightarrow \frac{3 x}{5} \sqrt{1-\frac{16 x^2}{25}}+\frac{4 x}{5} \sqrt{1-\frac{9 x^2}{25}}=x
\end{aligned}$
Either $x=0$
$\begin{aligned}
& \text {or } \frac{3 \sqrt{25-16 x^2}}{25}+\frac{4 \sqrt{25-9 x^2}}{25}=1 \\
& \Rightarrow 9\left(25-16 x^2\right)+16\left(25-9 x^2\right)+24 \sqrt{25-16 x^2}
\end{aligned}$
$\begin{aligned}
& \sqrt{25-9 x^2}=625 \\
& \Rightarrow 24 \sqrt{25-16 x^2} \sqrt{25-9 x^2}=288 x^2 \\
& \Rightarrow\left(25-16 x^2\right)\left(25-9 x^2\right)=144 x^4 \\
& \Rightarrow 625-625 x^2=0 \Rightarrow x= \pm 1
\end{aligned}$
So, sum of the values of '$x$'
$=0+1-1=0$
Hence, option (a) is correct.
Asked in: AP EAMCET 2019 (20 Apr Shift 2)
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