1) sin2x-корень из 3 3) sincx

Let's analyze each inequality step by step. ### 1) \( \sin(2x) < 0 \) The sine function is negative in the third and fourth quadrants. The general solutions for \( \sin(2x) < 0 \) occur in the intervals: \[ \frac{(2n+1)\pi}{2} < 2x < (2n+2)\pi \quad (n \in \mathbb{Z}) \] Dividing through by 2 gives: \[ \frac{(2n+1)\pi}{4} < x < (n+1)\pi \] This means the solution set for \( x \) can be written as: \[ x \in \left( \frac{(2n+1)\pi}{4}, (n+1)\pi \right), \quad n \in \mathbb{Z} \] ### 2) \( 2\cos(3x) > -\sqrt{3} \) First, we divide the entire inequality by 2: \[ \cos(3x) > -\frac{\sqrt{3}}{2} \] The cosine function is greater than \(-\frac{\sqrt{3}}{2}\) in the intervals: \[ 2k\pi - \frac{\pi}{6} < 3x < 2k\pi + \frac{\pi}{6} \quad (k \in \mathbb{Z}) \] Dividing by 3 gives: \[ \frac{2k\pi - \frac{\pi}{6}}{3} < x < \frac{2k\pi + \frac{\pi}{6}}{3} \] This leads to the solution set: \[ x \in \left( \frac{2k\pi}{3} - \frac{\pi}{18}, \frac{2k\pi}{3} + \frac{\pi}{18} \right), \quad k \in \mathbb{Z} \] ### 3) \( \sin(x) < \cos(x) \) To solve this inequality, we can rewrite it as: \[ \tan(x) < 1 \] This occurs in the intervals where the tangent function is less than 1. The general solution is given by: \[ x < \frac{\pi}{4} + n\pi \quad (n \in \mathbb{Z}) \] The intervals where \( \tan(x) < 1 \) are: \[ (x \in (n\pi, n\pi + \frac{\pi}{4})) \quad (n \in \mathbb{Z}) \] ### Summary of Solutions: 1. \( \sin(2x) < 0: \quad x \in \left( \frac{(2n+1)\pi}{4}, (n+1)\pi \right), n \in \mathbb{Z} \) 2. \( 2\cos(3x) > -\sqrt{3}: \quad x \in \left( \frac{2k\pi}{3} - \frac{\pi}{18}, \frac{2k\pi}{3} + \frac{\pi}{18} \right), k \in \mathbb{Z} \) 3. \( \sin(x) < \cos(x): \quad x \in (n\pi, n\pi + \frac{\pi}{4}), n \in \mathbb{Z} \) Make sure to apply the solutions appropriately depending on the specific domain required for your problems or exam.