Question

Ctg (π/4+x)<1 тригометрические неравенства​

Answer 1

Только перевёрнуто получилось:)

Answer 2

To solve the trigonometric inequality $\cot(\frac{\pi}{4}+x) < 1$, we need to find the range of values of $x$ that satisfy the inequality. Let's break it down step by step:

1. Start with the given inequality: $\cot(\frac{\pi}{4}+x) < 1$.
2. First, we need to find the cotangent of the angle $\frac{\pi}{4}+x$. Recall that $\cot(\theta) = \frac{1}{\tan(\theta)}$.
3. Since $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$, we can rewrite the inequality as $\cot(\frac{\pi}{4}+x) = \frac{1}{\tan(\frac{\pi}{4}+x)} = \frac{1}{\frac{\sin(\frac{\pi}{4}+x)}{\cos(\frac{\pi}{4}+x)}}$.
4. Next, we simplify the expression: $\cot(\frac{\pi}{4}+x) = \frac{\cos(\frac{\pi}{4}+x)}{\sin(\frac{\pi}{4}+x)}$.

Now, we have the inequality $\frac{\cos(\frac{\pi}{4}+x)}{\sin(\frac{\pi}{4}+x)} < 1$.

5. To further simplify, remember that $\cos(\frac{\pi}{4}) = \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$. Therefore, $\cos(\frac{\pi}{4}+x) = \sin(\frac{\pi}{4}+x) = \frac{1}{\sqrt{2}}$.

Now, the inequality becomes $\frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} < 1$.

6. The inequality now becomes $1 < 1$, which is not true for any value of $x$.

Therefore, the inequality $\cot(\frac{\pi}{4}+x) < 1$ has no solutions for $x$ in the real number domain.