Question

6cos^2x+cos^3x=-5cosx

Answer 1

Обозначим cosx=у
6у²+у³+5у=0
у(у²+6у+5)=0
1) у=0 
cosx=0
x=π/2 +πk, k∈Z
2) y²+6y+5=0
d=36-20=16
у1=(-6+4)/2=-1
х=π+2πn, n∈Z

y2=(-6-4)/2=-5 этот корень не подходит т.к.   -1≤cosx≤1

Answer 2

To solve the equation $6\cos^2(x) + \cos^3(x) = -5\cos(x)$, we can first rewrite it as a quadratic equation in terms of $\cos(x)$ and then solve for $\cos(x)$.

Let's make a substitution to simplify the equation:

Let $u = \cos(x)$. Then, we have:

$6u^2 + u^3 = -5u$

Now, the equation is a cubic equation in $u$. To solve for $u$, we can rearrange it to the form of a cubic equation:

$u^3 + 6u^2 + 5u = 0$

Now, we can factor out $u$ from the equation:

$u(u^2 + 6u + 5) = 0$

Next, we can factor the quadratic expression within the parentheses:

$u(u + 1)(u + 5) = 0$

Now, we have a product of three factors, and we can set each factor equal to zero and solve for $u$:

1. $u = 0$
2. $u + 1 = 0$ => $u = -1$
3. $u + 5 = 0$ => $u = -5$

Now, we have found the values of $u$, which represent the possible values of $\cos(x)$. To find the solutions for $x$, we can use the inverse cosine function:

1. For $u = 0$, $\cos(x) = 0$. So, $x = \cos^{-1}(0) = \frac{\pi}{2} + n\pi$, where $n$ is an integer.

2. For $u = -1$, $\cos(x) = -1$. So, $x = \cos^{-1}(-1) = \pi + n\pi$, where $n$ is an integer.

3. For $u = -5$, $\cos(x) = -5$. However, this value is not within the range of the cosine function, so there are no real solutions for this case.

So, the solutions for the equation $6\cos^2(x) + \cos^3(x) = -5\cos(x)$ are:

1. $x = \frac{\pi}{2} + n\pi$
2. $x = \pi + n\pi$