Question

8 cos 2 x +14 cos X -9 =0

Answer 1

8cos2x + 14cosx - 9 = 0
8*(2cos^2x - 1) + 14cosx - 9 = 0
16cos^2x - 8 + 14cosx - 9 = 0
16cos^2x + 14cosx - 17 = 0
D = 196 + 4*16*17 = 196 + 1088 = 1284
корень из числа 1284 не извлекается. Надо проверить условие.

Answer 2

To solve the equation $8\cos^2(x) + 14\cos(x) - 9 = 0$, we can use a substitution to simplify it. Let $u = \cos(x)$, then the equation becomes:

$8u^2 + 14u - 9 = 0$

Now, we can solve this quadratic equation for $u$. You can use the quadratic formula, which states that for an equation in the form $ax^2 + bx + c = 0$, the solutions for $x$ are given by:

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

In this case, $a = 8$, $b = 14$, and $c = -9$. Plugging these values into the quadratic formula:

$u = \frac{-14 \pm \sqrt{14^2 - 4(8)(-9)}}{2(8)}$

Now, let's calculate the values of $u$:

$u = \frac{-14 \pm \sqrt{196 + 288}}{16}$ $u = \frac{-14 \pm \sqrt{484}}{16}$ $u = \frac{-14 \pm 22}{16}$

Now, we have two possible solutions for $u$:

1. $u_1 = \frac{-14 + 22}{16} = \frac{8}{16} = \frac{1}{2}$
2. $u_2 = \frac{-14 - 22}{16} = \frac{-36}{16} = -\frac{9}{4}$

Now that we have the solutions for $u$, we can find the solutions for $x$ by using the inverse cosine function:

1. For $u = \frac{1}{2}$: $x_1 = \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} + 2\pi n$ (where $n$ is an integer)

2. For $u = -\frac{9}{4}$: Since $-1 \leq \cos(x) \leq 1$, there are no real solutions for $x$ in this case.

So, the equation $8\cos^2(x) + 14\cos(x) - 9 = 0$ has one real solution:

$x = \frac{\pi}{3} + 2\pi n$