Question

5 tg^2(х-П)+12 tg(-x+П)=0

Answer 1

Ответ:

Ответ смотри во вложении

Answer 2

This equation seems like a trigonometric one. To solve it, you can use some trigonometric identities and properties. Let's break it down step by step.

The given equation is $5\tan^2(x-\pi) + 12\tan(-x+\pi) = 0$.

First, let's use the fact that $\tan(\theta) = -\tan(\theta + \pi)$.

So, $5\tan^2(x-\pi) + 12\tan(-x+\pi)$ can be written as $5\tan^2(x-\pi) - 12\tan(x)$.

Now, we can use the identity $\tan^2(\theta) + 1 = \sec^2(\theta)$.

Replace $\tan^2(x-\pi)$ with $\sec^2(x-\pi)$: $5\sec^2(x-\pi) - 12\tan(x) = 0$.

Now, you can solve this equation for $x$. If it's still a bit tricky, let me know where you get stuck, and I can guide you further!