Find the number of positive solutions to
\[\sin (\arccos( \tan (\arcsin x) ) ) ) = x.\]
Let $\theta = \arcsin x,$ so $0 < \theta < \frac{\pi}{2}$ and $\sin \theta = x.$  Then
\[\cos \theta = \sqrt{1 - x^2},\]so
\[\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{x}{\sqrt{1 - x^2}}.\]Thus,
\[\sin (\arccos (\tan (\arcsin x) ) ) ) = \sin \left( \arccos \frac{x}{\sqrt{1 - x^2}} \right).\]Let $\psi = \arccos \frac{x}{\sqrt{1 - x^2}},$ so $0 < \psi < \frac{\pi}{2}$ and $\cos \psi = \frac{x}{\sqrt{1 - x^2}}.$  Then
\[\sin \psi = \sqrt{1 - \cos^2 \psi} = \sqrt{1 - \frac{x^2}{1 - x^2}} = \sqrt{\frac{1 - 2x^2}{1 - x^2}} = x.\]Squaring both sides, we get
\[\frac{1 - 2x^2}{1 - x^2} = x^2.\]Then $1 - 2x^2 = x^2 - x^4,$ so $x^4 - 3x^2 + 1 = 0.$  By the quadratic formula,
\[x^2 = \frac{3 \pm \sqrt{5}}{2}.\]The positive solutions are then $\sqrt{\frac{3 + \sqrt{5}}{2}}$ and $\sqrt{\frac{3 - \sqrt{5}}{2}}.$  However, $\arcsin x$ is defined only for $-1 \le x \le 1,$ so there is only $\boxed{1}$ positive solution, namely
\[x = \sqrt{\frac{3 - \sqrt{5}}{2}}.\]
1
Level 2
Precalculus
Problem: Find the number of positive solutions to
\[\sin (\arccos( \tan (\arcsin x) ) ) ) = x.\]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the equation, we first need to understand the functions involved. The function \(\arcsin x\) is defined for \(x \in [0, 1]\) and returns values in \([0, \frac{\pi}{2}]\). Therefore, \(x\) must be in the interval \((0, 1]\) for the equation to have positive solutions.

Next, we evaluate \(\tan(\arcsin x)\). The output of \(\arcsin x\) is an angle \(\theta\) such that \(\sin \theta = x\). Using the identity \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) and knowing that \(\cos \theta = \sqrt{1 - \sin^2 \theta}\), we find:

\[
\tan(\arcsin x) = \frac{x}{\sqrt{1 - x^2}}.
\]

Now, we substitute this into \(\arccos\):

\[
\arccos\left(\tan(\arcsin x)\right) = \arccos\left(\frac{x}{\sqrt{1 - x^2}}\right).
\]

Next, we need to evaluate \(\sin(\arccos(y))\), which is given by the identity \(\sin(\arccos(y)) = \sqrt{1 - y^2}\). Thus, we have:

\[
\sin\left(\arccos\left(\frac{x}{\sqrt{1 - x^2}}\right)\right) = \sqrt{1 - \left(\frac{x}{\sqrt{1 - x^2}}\right)^2}.
\]

Calculating this gives:

\[
\sqrt{1 - \frac{x^2}{1 - x^2}} = \sqrt{\frac{1 - x^2 - x^2}{1 - x^2}} = \sqrt{\frac{1 - 2x^2}{1 - x^2}}.
\]

Setting this equal to \(x\) leads to the equation:

\[
\sqrt{\frac{1 - 2x^2}{1 - x^2}} = x.
\]

Squaring both sides results in:

\[
\frac{1 - 2x^2}{1 - x^2} = x^2.
\]

Cross-multiplying gives:

\[
1 - 2x^2 = x^2(1 - x^2) \implies 1 - 2x^2 = x^2 - x^4.
\]

Rearranging terms leads to:

\[
x^4 - 3x^2 + 1 = 0.
\]

Letting \(y = x^2\), we convert this to a quadratic equation:

\[
y^2 - 3y + 1 = 0.
\]

Using the quadratic formula, we find:

\[
y = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2}.
\]

Since \(y = x^2\) must be positive, we take the positive root:

\[
y = \frac{3 - \sqrt{5}}{2}.
\]

Calculating \(x\) gives:

\[
x = \sqrt{\frac{3 - \sqrt{5}}{2}}.
\]

We need to check if this value lies within the interval \((0, 1]\). Since \(3 - \sqrt{5} < 3\), it follows that \(\frac{3 - \sqrt{5}}{2} < 1\), confirming \(x\) is a valid solution.

Finally, we verify if there are any other solutions. The original equation is derived from continuous functions, and the transformations applied are monotonic in the specified intervals. Thus, the only positive solution is the one we found.

The final answer is \(1\).<|END_OF_LATENT|> The final answer is $1$.