For $x \in(-1,1]$, the number of solutions of the equation $\sin ^{-1} x=2 \tan ^{-1} x$ is equal to __________.

<p>We&#39;re given the equation $\sin^{-1}x = 2\tan^{-1}x$ for $x$ in the interval $(-1, 1]$. We want to find the number of solutions.</p> <p>Step 1: Apply the sine and tangent functions to both sides :</p> <p>We can rewrite the equation by applying the sine function to both sides :</p> <p>$\sin(\sin^{-1}x) = \sin(2\tan^{-1}x).$</p> <p>This simplifies to:</p> <p>$x = \sin(2\tan^{-1}x).$</p> <p>Step 2: Use the double-angle identity for sine :</p> <p>Recall that $\sin(2y) = 2\sin(y)\cos(y)$. Applying this identity to the right-hand side gives :</p> <p>$x = 2\sin(\tan^{-1}x)\cos(\tan^{-1}x).$</p> <p>Step 3: Use the identities for sine and cosine of an inverse tangent :</p> <p>Recall that $\sin(\tan^{-1}x) = \frac{x}{\sqrt{1 + x^2}}$ and $\cos(\tan^{-1}x) = \frac{1}{\sqrt{1 + x^2}}$. Substituting these into the equation gives :</p> <p>$x = 2 \cdot \frac{x}{\sqrt{1 + x^2}} \cdot \frac{1}{\sqrt{1 + x^2}}.$</p> <p>This simplifies to :</p> <p>$x = \frac{2x}{1 + x^2}.$</p> <p>Step 4: Solve for $x$ :</p> <p>We have :</p> <p>$x = \frac{2x}{1 + x^2}.$</p> <p>Cross-multiplying gives :</p> <p>$x(1 + x^2) = 2x.$</p> <p>This simplifies to :</p> <p>$x^3 + x - 2x = 0.$</p> <p>Rearranging terms gives :</p> <p>$x^3 - x = 0.$</p> <p>This factors to:</p> <p>$x(x^2 - 1) = 0.$</p> <p>Setting each factor equal to zero gives the solutions $x = 0$, $x = -1$, and $x = 1$.</p> <p>However, we are given that $x \in (-1, 1]$. Therefore, the only solutions in this interval are $x = 0$ and $x = 1$.</p> <p>So there are 2 solutions to the equation $\sin ^{-1} x=2 \tan ^{-1} x$ in the interval $x \in(-1,1]$.</p>