Question

Решите неравеснтво: arcsinx<arccosx

Answer 1

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Answer 2

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Answer 3

To solve the inequality $\arcsin(x) < \arccos(x)$, we need to find the values of $x$ that satisfy the inequality.

First, let's recall the properties of the arcsin and arccos functions:

1. $\arcsin(x)$ is defined for $x$ in the range $[-1, 1]$ and returns angles in the range $[- \frac{\pi}{2}, \frac{\pi}{2}]$.
2. $\arccos(x)$ is defined for $x$ in the range $[-1, 1]$ and returns angles in the range $[0, \pi]$.

Now, to find the values of $x$ that satisfy the inequality, we'll consider two cases:

Case 1: $x \geq 0$

For $x$ in the range $[0, 1]$, both $\arcsin(x)$ and $\arccos(x)$ are in the range $[0, \frac{\pi}{2}]$.

In this case, the inequality $\arcsin(x) < \arccos(x)$ will be satisfied for all $x$ in the range $[0, 1)$, excluding $x = 1$.

Case 2: $x < 0$

For $x$ in the range $[-1, 0)$, $\arcsin(x)$ is in the range $(-\frac{\pi}{2}, 0]$, and $\arccos(x)$ is in the range $[\frac{\pi}{2}, \pi]$.

In this case, the inequality $\arcsin(x) < \arccos(x)$ will not be satisfied for any $x$ in the range $[-1, 0)$.

Putting both cases together, we find that the inequality $\arcsin(x) < \arccos(x)$ is satisfied for all $x$ in the range $[0, 1)$.

In interval notation, the solution to the inequality is $x \in [0, 1)$.