Question

Given that the inverse trigonometric functions take principal values only.

Then, the number of real values of $x$ which satisfy ${\sin }^{-1}\left(\frac{3x}{5}\right)+{\sin }^{-1}\left(\frac{4x}{5}\right)={\sin }^{-1}x$

Answer 1

Finding the number of real values of x:

Given equation is ${\sin }^{-1}\left(\frac{3x}{5}\right)+{\sin }^{-1}\left(\frac{4x}{5}\right)={\sin }^{-1}x$

Taking sin on both sides,$\left(\frac{3x}{5}\right)\sqrt{[1–(\frac{16x^2}{25}\left)\right]}+\left(\frac{4x}{5}\right)\sqrt{[1–(\frac{9x^2}{25}\left)\right]}=x$

$$\begin{gathered} 3x\sqrt{(25–16x^2)}=25x–4x\sqrt{(25–9x^2)} \\ \Rightarrow x=0 \\ or3\sqrt{(25–16x^2)}=25–4\sqrt{(25–9x^2)} \\ \Rightarrow 9(25–16x^2)=625–200\sqrt{(25–9x^2)}+16(25–9x^2) \\ \Rightarrow 200\sqrt{(25–9x^2)}=800 \\ \Rightarrow \sqrt{(25–9x^2)}=4 \\ \Rightarrow x^2=1 \\ \Rightarrow x=\pm 1 \end{gathered}$$

Total number of solutions $=3$

Hence, the number of real values of $x$ is $3$