Question

The equation ${\sin }^{-1}\left(\frac{5x}{13}\right)+{\sin }^{-1}\left(\frac{12x}{13}\right)={\sin }^{-1}x$ has

• A. no root
• B. two roots
• C. three roots
• D. its sum of root is zero

Answer 1

Answer: (C), (D)

The correct options are
C three roots
D its sum of root is zero

If we carefully observe, then we can see that x=0 is one of the solution to the above question.
Now
we know that 5,12,13 are Pythagorean triplet.
Hence for some angle (say $\theta$), in a right angled triangle with sides 5,12,13 units,
$sin\left(\theta \right)=\frac{5}{13}$
$cos\left(\theta \right)=sin({90}^0-\theta )=\frac{12}{13}$
Hence $si{n}^{-1}\left(sin\right(\theta \left)\right)+si{n}^{-1}\left(cos\right(\theta \left)\right)$
$=si{n}^{-1}\left(\frac{5}{13}\right)+si{n}^{-1}\left(\frac{12}{13}\right)$
$=\theta +{90}^0-\theta$
$={90}^0$
$=si{n}^{-1}\left(1\right)$
Hence we get another root, for the above equation, i.e x=1.
Now if we take the $\theta$ as $-\theta$, that is in clockwise sense, then we get another value of x as x=-1.
Hence total three roots will be there,
-1,0,1.
Sum of the roots is 0.