Question

Sin^2(x)+sin(x)-2=0, [0;-п] ​

Answer 1

Ответ:

Пошаговое объяснение:

sin²(x)+sin(x)-2=0, [0;-п] ​   обозначим sin(x)=к

к²+к-2=0  

по теореме Виета произведение корней равно -2 а сумма равна -1

значит крни  к=-2 и к=1

sin(x)=1 ,  второй вариант не существует. sin(x) ≠ -2     -1≤sin(x)≤1

sin(x)=1  х=п/2

Answer 2

To solve the equation $\sin^2(x) + \sin(x) - 2 = 0$ on the interval $[0, \pi]$, you can use a substitution. Let $u = \sin(x)$, so the equation becomes a quadratic equation in $u$:

$u^2 + u - 2 = 0$

Now, you can solve this quadratic equation for $u$. The solutions will be values of $\sin(x)$, and then you can find the corresponding values of $x$ within the given interval.

Let's solve for $u$:

$u^2 + u - 2 = 0$

Factor the quadratic equation:

$(u + 2)(u - 1) = 0$

Now, set each factor equal to zero:

1. $u + 2 = 0$ $u = -2$

2. $u - 1 = 0$ $u = 1$

Now, we have two potential solutions for $u$, which correspond to possible values of $\sin(x)$. We need to find the corresponding values of $x$ within the interval $[0, \pi]$.

For $u = -2$: $\sin(x) = -2$

However, there are no real solutions to this equation because the sine function is bounded between -1 and 1, so $\sin(x)$ cannot equal -2.

For $u = 1$: $\sin(x) = 1$

This equation has solutions within the interval $[0, \pi]$, specifically $x = \frac{\pi}{2}$.

So, the only solution to the equation $\sin^2(x) + \sin(x) - 2 = 0$ on the interval $[0, \pi]$ is $x = \frac{\pi}{2}$.