Question

Sin^2 x + 2 cos x + 2 = 0​

Answer 1

$\sin^{2}x + 2\cos x + 2 = 0$

$1 - \cos^{2}x + 2 \cos x + 2 = 0$

$-\cos^{2}x + 2\cos x + 3 = 0$

$\cos^{2}x - 2\cos x - 3 = 0$

Замена: $\cos x = t, \ t \in [-1; \ 1]$

$t^{2} - 2t - 3 = 0$

$t_{1}=-1$

$t_{2} = 3 \notin [-1; \ 1]$

Обратная замена:

$\cos x = -1$

$x = \pi + 2\pi n, \ n \in Z$

Ответ: $x = \pi + 2\pi n, \ n \in Z$

Answer 2

To solve the equation $\sin^2(x) + 2 \cos(x) + 2 = 0$, we can start by making use of the trigonometric identity $\sin^2(x) + \cos^2(x) = 1$. Rearrange the given equation to isolate the term involving $\sin^2(x)$:

$\sin^2(x) = -2 \cos(x) - 2$

Now, we can replace $\sin^2(x)$ using the identity:

$1 - \cos^2(x) = -2 \cos(x) - 2$

Rearrange to get a quadratic equation:

$\cos^2(x) + 2\cos(x) + 3 = 0$

Now we have a quadratic equation in terms of $\cos(x)$. We can solve this quadratic equation using the quadratic formula:

The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

In our case, $a = 1$, $b = 2$, and $c = 3$.

$\cos(x) = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1}$

$\cos(x) = \frac{-2 \pm \sqrt{4 - 12}}{2}$

$\cos(x) = \frac{-2 \pm \sqrt{-8}}{2}$

Since the square root of a negative number is not a real number, there are no real solutions for $\cos(x)$. Therefore, the original equation $\sin^2(x) + 2 \cos(x) + 2 = 0$ has no real solutions for $x$.