Question

5 sin 2 x + 5 cos x +5 = 0

Answer 1

Ответ:

П+2Пn

Пошаговое объяснение:

Все написанно на листочке

Answer 2

To solve the trigonometric equation $5 \sin^2(x) + 5 \cos(x) + 5 = 0$, we can use trigonometric identities to express everything in terms of either $\sin$ or $\cos$. Let's start by using the Pythagorean identity: $\sin^2(x) + \cos^2(x) = 1$.

Since $5 \sin^2(x) + 5 \cos(x) + 5 = 0$, we can divide the entire equation by 5 to simplify it:

$\sin^2(x) + \cos(x) + 1 = 0$

Now, using the Pythagorean identity $\sin^2(x) = 1 - \cos^2(x)$, we can substitute this into the equation:

$1 - \cos^2(x) + \cos(x) + 1 = 0$

Rearrange the terms:

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

Now, this is a quadratic equation in terms of $\cos(x)$. Let's solve it by factoring:

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

$-(\cos^2(x) - \cos(x) - 2) = 0$

$-(\cos(x) - 2)(\cos(x) + 1) = 0$

Now, set each factor equal to zero and solve for $\cos(x)$:

1. $\cos(x) - 2 = 0$ $\cos(x) = 2$ (Not a valid solution since the range of $\cos$ is $[-1, 1]$).

2. $\cos(x) + 1 = 0$ $\cos(x) = -1$

Now, we have one valid solution for $\cos(x) = -1$. To find the corresponding values of $x$, we can use the inverse cosine function:

$x = \cos^{-1}(-1)$

The principal value of $\cos^{-1}(-1)$ is $\pi$, but we should also consider other solutions since the cosine function has a periodic nature. In general, the solutions can be expressed as:

$x = \pi + 2n\pi$, where $n$ is an integer.

So, the general solution to the equation $5 \sin^2(x) + 5 \cos(x) + 5 = 0$ is:

$x = \pi + 2n\pi$, where $n$ is an integer.