Question

2sin(2x-п/3)= минус 

Answer 1

$2\sin ( 2x-\frac{\pi}{3} )=- \sqrt{3} \\ \sin ( 2x-\frac{\pi}{3} )=- \frac{ \sqrt{3} }{2} \\ 2x-\frac{\pi}{3}=(-1)^k^+^1\cdot \frac{\pi}{3} + \pi k,k \in Z \\ 2x=(-1)^k^+^1\cdot \frac{\pi}{3} +\frac{\pi}{3}+ \pi k,k \in Z \\ x=(-1)^k^+^1\cdot \frac{\pi}{6}+\frac{\pi}{6}+\frac{\pi k}{2}, k \in Z$

Answer 2

To solve the equation 2sin(2x - π/3) = -1, you can follow these steps:

1. First, isolate the sine function by dividing both sides by 2:

   sin(2x - π/3) = -1/2

2. Next, you want to find the angles where sine equals -1/2. In the unit circle, sine is equal to -1/2 at two specific angles: π/6 and 5π/6.

3. Set up two equations for the angles:

   2x - π/3 = π/6 and 2x - π/3 = 5π/6

4. Solve each equation for x:

   For the first equation: 2x - π/3 = π/6 2x = π/6 + π/3 2x = π/6 + 2π/6 2x = 3π/6 x = 3π/12 x = π/4

   For the second equation: 2x - π/3 = 5π/6 2x = 5π/6 + π/3 2x = 5π/6 + 2π/6 2x = 7π/6 x = 7π/12

So, the solutions for the equation 2sin(2x - π/3) = -1 are x = π/4 and x = 7π/12.