Question

10cos^2 x-19 sin x-17=0

Answer 1

Решение смотри на фотографии

Answer 2

To solve the trigonometric equation $10\cos^2 x - 19\sin x - 17 = 0$, we'll use the trigonometric identities $\cos^2 x = 1 - \sin^2 x$ and $\sin^2 x + \cos^2 x = 1$ to rewrite the equation in terms of a single trigonometric function. Let's proceed step by step:

1. Replace $\cos^2 x$ with $1 - \sin^2 x$: $10(1 - \sin^2 x) - 19\sin x - 17 = 0$

2. Distribute and simplify: $10 - 10\sin^2 x - 19\sin x - 17 = 0$

3. Move all terms to one side of the equation: $10\sin^2 x + 19\sin x + 7 = 0$

Now, we have a quadratic equation in terms of $\sin x$. Let's solve it using the quadratic formula:

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

For our equation, $10\sin^2 x + 19\sin x + 7 = 0$, we have $a = 10$, $b = 19$, and $c = 7$.

Applying the quadratic formula:

$\sin x = \frac{-19 \pm \sqrt{19^2 - 4 \cdot 10 \cdot 7}}{2 \cdot 10}$

$\sin x = \frac{-19 \pm \sqrt{361 - 280}}{20}$

$\sin x = \frac{-19 \pm \sqrt{81}}{20}$

$\sin x = \frac{-19 \pm 9}{20}$

Now, we have two possible solutions for $\sin x$:

1. $\sin x = \frac{-19 + 9}{20} = \frac{-10}{20} = -\frac{1}{2}$
2. $\sin x = \frac{-19 - 9}{20} = \frac{-28}{20} = -\frac{7}{5}$

However, $\sin x$ cannot be greater than 1 or less than -1 (as it represents the sine of an angle). Therefore, the second solution, $\sin x = -\frac{7}{5}$, is not valid.

So, the valid solution is:

$\sin x = -\frac{1}{2}$

Now, we need to find the values of $x$ that satisfy this equation. From the unit circle or trigonometric tables, we know that $\sin x = -\frac{1}{2}$ at two specific angles: $x = -\frac{\pi}{6}$ and $x = -\frac{5\pi}{6}$. However, sine is a periodic function, and we can find an infinite number of solutions by adding multiples of $2\pi$ to these angles.

So, the general solutions for $x$ are:

$x = -\frac{\pi}{6} + 2n\pi$ and $x = -\frac{5\pi}{6} + 2n\pi$, where $n$ is an integer.

These are the solutions to the trigonometric equation $10\cos^2 x - 19\sin x - 17 = 0$