Question

Tg²3x-2tg²3x+tg3x-2=0​

Answer 1

Ответ:

Ответ смотри во вложении

Answer 2

To solve the equation $\tan^2(3x) - 2\tan^2(3x) + \tan(3x) - 2 = 0$, we can simplify it by factoring and then solve for $x$. Let's break it down step by step:

1. Start with the equation: $\tan^2(3x) - 2\tan^2(3x) + \tan(3x) - 2 = 0$

2. Factor out a common factor of $\tan^2(3x)$: $\tan^2(3x) \left(1 - 2 + \frac{\tan(3x)}{\tan^2(3x)} - \frac{2}{\tan^2(3x)}\right) = 0$

3. Simplify the expression inside the parentheses: $\tan^2(3x) (-1 + \frac{\tan(3x)}{\tan^2(3x)} - \frac{2}{\tan^2(3x)}) = 0$

4. Now, let's deal with the fractions. Remember that $\tan(x) = \frac{\sin(x)}{\cos(x)}$: $\tan^2(3x) \left(-1 + \frac{\frac{\sin(3x)}{\cos(3x)}}{\frac{\sin^2(3x)}{\cos^2(3x)}} - \frac{2}{\frac{\sin^2(3x)}{\cos^2(3x)}}\right) = 0$

5. Simplify further by multiplying through by $\cos^2(3x)$ to get rid of the fractions: $\tan^2(3x) \left(-\cos^2(3x) + \frac{\sin(3x)}{\cos(3x)} - 2\cos^2(3x)\right) = 0$

6. Now, use the trigonometric identity $\tan(x) = \frac{\sin(x)}{\cos(x)}$: $\tan^2(3x) \left(-\cos^2(3x) + \tan(3x) - 2\cos^2(3x)\right) = 0$

7. Distribute the $\tan^2(3x)$ term: $-\tan^2(3x)\cos^2(3x) + \tan^3(3x) - 2\tan^2(3x)\cos^2(3x) = 0$

8. Combine like terms: $-3\tan^2(3x)\cos^2(3x) + \tan^3(3x) = 0$

9. Factor out a common factor of $\tan^2(3x)$: $\tan^2(3x)(\tan(3x) - 3\cos^2(3x)) = 0$

Now, you have two factors to consider:

1. $\tan^2(3x) = 0$: This implies $\tan(3x) = 0$. Solving for $x$: $3x = k\pi$, where $k$ is an integer. $x = \frac{k\pi}{3}$

2. $\tan(3x) - 3\cos^2(3x) = 0$