Question

Cos2x<0 Pomogite......

Answer 1

$Cos2x$

Answer 2

$\cos(2x) < 0 \\ - \frac{3\pi}{2} + 2\pi \: n < 2x < - \frac{\pi}{2} + 2\pi \: n \\ - \frac{3\pi}{4} + \pi \: n < x < - \frac{\pi}{4} + \pi \: n$

Answer 3

To solve the inequality cos(2x) < 0, we can follow these steps:

1. Find the critical points where cos(2x) = 0 or undefined. This happens when 2x = (2n+1)π/2, where n is an integer. Therefore, x = (2n+1)π/4.

2. Divide the number line into intervals between these critical points.

3. Test each interval by plugging in a test point into the inequality cos(2x) < 0. For example, if we test x = 0 in the interval (-π/4, π/4), we get cos(2x) = cos(0) = 1 > 0, so this interval is not a solution.

4. Determine the solution set by combining the intervals where the inequality is true.

Using these steps, we can find the solution set for cos(2x) < 0:

• The critical points are x = (2n+1)π/4 for all integers n.
• The intervals are (-π/2, -3π/4), (-π/4, π/4), and (3π/4, π/2).
• Testing a point in each interval, we find that cos(2x) < 0 in (-π/2, -3π/4) and (3π/4, π/2).
• Therefore, the solution set is x < -3π/8 or x > π/8.

Hence, the solution to the inequality cos(2x) < 0 is x < -3π/8 or x > π/8.