For [x (-1,1]], the number of solutions of the equation [ ^-1 x=2 ^-1 x] is equal to _____…

Question

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

Answer: 2

Accepted Answer

We're given the equation ${\sin }^{-1}x=2{	an }^{-1}x$ for $x$ in the interval $(-1,1]$. We want to find the number of solutions.

Apply the sine and tangent functions to both sides : 1: Apply the sine and tangent functions to both sides :

We can rewrite the equation by applying the sine function to both sides :

$\sin ({\sin }^{-1}x)=\sin (2{	an }^{-1}x).$

This simplifies to:

$x=\sin (2{	an }^{-1}x).$

Use the double-angle identity for sine : 2: Use the double-angle identity for sine :

Recall that $\sin (2y)=2\sin (y)\cos (y)$. Applying this identity to the right-hand side gives :

$x=2\sin ({	an }^{-1}x)\cos ({	an }^{-1}x).$

Use the identities for sine and cosine of an inverse tangent : 3: Use the identities for sine and cosine of an inverse tangent :

Recall that $\sin ({	an }^{-1}x)=\frac{x}{\sqrt{1+{x}^{2}}}$ and $\cos ({	an }^{-1}x)=\frac{1}{\sqrt{1+{x}^{2}}}$. Substituting these into the equation gives :

$x=2\cdot \frac{x}{\sqrt{1+{x}^{2}}}\cdot \frac{1}{\sqrt{1+{x}^{2}}}.$

This simplifies to :

$x=\frac{2x}{1+{x}^{2}}.$

Solve for $x$ : 4: Solve for $x$ :

We have :

$x=\frac{2x}{1+{x}^{2}}.$

Cross-multiplying gives :

$x(1+{x}^{2})=2x.$

This simplifies to :

${x}^{3}+x-2x=0.$

Rearranging terms gives :

${x}^{3}-x=0.$

This factors to:

$x({x}^{2}-1)=0.$

Setting each factor equal to zero gives the solutions $x=0$, $x=-1$, and $x=1$.

However, we are given that $x\in (-1,1]$. Therefore, the only solutions in this interval are $x=0$ and $x=1$.

So there are 2 solutions to the equation ${\sin }^{-1}x=2{	an }^{-1}x$ in the interval $x\in (-1,1]$.

Solutions · 1