Question

2arcsin^2x < 3arcsinx

Answer 1

Ответ:

(0; sin 1,5)

Объяснение:

2arcsin^2x < 3arcsinx

$2 \arcsin^2{x} < 3 \arcsin{x}$

По свойствам арксинуса, ОДЗ:

$x \in[{-1}{;} \: 1] \: = > \: \: \arcsin{x} \in[{- \frac{\pi}{2} }{;} \: \frac{\pi}{2} ]$

Замена переменной:

$t= \arcsin{x}; \:\: t\in[{- \frac{\pi}{2} }{;} \: \frac{\pi}{2} ] \\ 2 {t}^{2} < 3t \: < = > \: 2 {t}^{2} - 3t < 0 \\ 2t(t - 1.5) < 0 \\ ^{ \: + + \: \: \: \: \: \: - - \: \: \: \: \: \: + + }_ { - - - o - - - - o- - - > } \\ {}^{0 \: \: \: \: \: \: \: \: \: 1.5 \: \: } \\t \in ({0}{;} \: 1.5) \: < = > \arcsin{x} \in ({0}{;} \: 1.5)$

Очевидно, что:

$3 < \pi \: => \: 1.5$

а значит весь отрезок входит в ОДЗ

т.к. функция арксинуса строго возрастает на всей обл. определения, а из определения арксинуса известно что

$\small{x{= }\sin({\arcsin {x}}); \: x{ \in}[{-1}{;} \: 1]; \:\arcsin {x}{\in}[{- \frac{\pi}{2} }{;} \: \frac{\pi}{2} ] }$

получаем:

$0< \arcsin{x}$

Answer 2

To solve the inequality $2\arcsin^2(x) < 3\arcsin(x)$, we can first rewrite it in terms of a single variable, typically sine or cosine. To do this, we can use the relationship between the inverse sine function ($\arcsin$) and the sine function ($\sin$).

Let $y = \arcsin(x)$. This means $\sin(y) = x$.

Now, we can rewrite the inequality as:

$2y^2 < 3y$

To solve this quadratic inequality, let's first move all the terms to one side of the inequality:

$2y^2 - 3y < 0$

Now, factor the left side:

$y(2y - 3) < 0$

This inequality holds for the values of $y$ where one of the factors is negative and the other is positive. So, we need to find the intervals where $2y - 3 < 0$ and $y > 0$ (because $\arcsin(x)$ is only defined for $0 \leq x \leq 1$).

1. For $2y - 3 < 0$, we have $y < \frac{3}{2}$.
2. For $y > 0$, we know that $0 \leq y \leq \frac{\pi}{2}$ because the range of $\arcsin(x)$ is $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$.

So, the solution for the inequality $2\arcsin^2(x) < 3\arcsin(x)$ is:

$0 < y < \frac{3}{2}$

However, we need to convert this solution back to the original variable x\ by using the relationship \(\sin(y) = x. Since $\sin(0) = 0$ and $\sin\left(\frac{\pi}{2}\right) = 1$, the solution for $x$ is:

$0 < x < 1$

So, the solution to the inequality $2\arcsin^2(x) < 3\arcsin(x)$ is $0 < x < 1$.