Question

Cos^2x+sinx=-1 хϵ[0;2pi]

Answer 1

$no numeric noise key 1040^no numeric noise key 1039 x + sinx 1 \\ - 1038^no numeric noise key 1037 0 1036^no numeric noise key 1035 2 =0$

По теореме Виета

$sinx = 2 \\ \: \: sinx = - 1$

sinx = 2 не подходит

$sinx = - 1 \\ x = - \frac{\pi}{2} + 2\pi\:n$

С помощью двойного неравенства отберём корни на отрезке [0; 2π]

$0 \leqslant \ - \frac{ \pi }{2} + 2\pi \: n \leqslant 2\pi \\ 0 \leqslant 2n - 0.5 \leqslant 2 \\ 0.5 \leqslant 2n \leqslant 2.5 \\ 0.25 \leqslant n \leqslant 1.25$

Т.к. n є Z, n = 1

$- \frac{\pi}{2} + 2\pi \times 1 = \frac{3\pi}{2}$

Ответ: 3π/2

Answer 2

The equation you've given, cos^2(x) + sin(x) = -1, does not have any solutions within the interval [0, 2π].

Let's break it down:

1. First, note that both sin(x) and cos(x) are bounded between -1 and 1. Therefore, cos^2(x) is always between 0 and 1. So, cos^2(x) + sin(x) is always greater than or equal to 0.

2. Since the sum of two non-negative numbers cannot be equal to -1, there are no solutions to this equation in the given interval.

In other words, within the interval [0, 2π], there are no values of x that satisfy the equation cos^2(x) + sin(x) = -1.