Question

Sin^2 x +11 cos x +41 =0

Answer 1

1-сos²x+11cosx+41=0
cosx=a
a²-11a-42=0
a1+a2=11 U a1*a2=-42
a1=-3 ⇒cosx=-3<-1 нет решения
a2=14⇒cosx=14>1-нет решения

Answer 2

To solve the trigonometric equation $\sin^2(x) + 11\cos(x) + 41 = 0$, we can use the trigonometric identity $\sin^2(x) + \cos^2(x) = 1$ to rewrite the equation in terms of a single trigonometric function. First, subtract 1 from both sides:

$$\sin^2(x) - 1 + 11\cos(x) + 41 = 0$$

Now, we can use the identity $\sin^2(x) = 1 - \cos^2(x)$ to substitute for $\sin^2(x)$:

$$(1 - \cos^2(x)) - 1 + 11\cos(x) + 41 = 0$$

Simplify further:

$$- \cos^2(x) - 11\cos(x) + 41 = 0$$

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

$$\cos(x) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

In this case, $a = -1$, $b = -11$, and $c = 41$. Plugging these values into the quadratic formula:

$$\cos(x) = \frac{11 \pm \sqrt{(-11)^2 - 4(-1)(41)}}{2(-1)}$$

Now, calculate the discriminant:

$$\cos(x) = \frac{11 \pm \sqrt{121 + 164}}{-2}$$

$$\cos(x) = \frac{11 \pm \sqrt{285}}{-2}$$

Now, you have two possible solutions for $\cos(x)$:

1. $\cos(x) = \frac{11 + \sqrt{285}}{-2}$
2. $\cos(x) = \frac{11 - \sqrt{285}}{-2}$

To find the corresponding values of $x$, you can take the arccosine (inverse cosine) of these values:

1. x = \arccos\left(\frac{11 + \sqrt{285}}{-2}\
2. x = \arccos\left(\frac{11 - \sqrt{285}}{-2}\

These will give you the solutions for $x$ in the range of $[0, 2\pi]$, but remember that cosine is a periodic function, so you can find solutions in other intervals by adding or subtracting multiples of $2\pi$.