2cos x ≤ -√2
To solve the inequality \( 2\cos x \leq -\sqrt{2} \), we can start by isolating \( \cos x \):
\[ \cos x \leq -\frac{\sqrt{2}}{2} \]
Now, we need to find the values of \( x \) for which this inequality holds. The cosine function is negative in the second and third quadrants. Specifically:
The cosine value \( -\frac{\sqrt{2}}{2} \) corresponds to angles where \( \cos x = -\frac{\sqrt{2}}{2} \). The angles in the range \( [0, 2\pi) \) that satisfy this are:
\[ x = \frac{3\pi}{4}, \, \frac{5\pi}{4} \]
The cosine function is less than \( -\frac{\sqrt{2}}{2} \) between these two angles:
\[ x \in \left( \frac{3\pi}{4}, \frac{5\pi}{4} \right) \]
Because cosine is periodic with period \( 2\pi \), we can express the general solutions for \( x \) as:
\[ x \in \left( \frac{3\pi}{4} + 2k\pi, \frac{5\pi}{4} + 2k\pi \right) \quad \text{for any integer } k \]
In summary, the solution to the inequality \( 2\cos x \leq -\sqrt{2} \) is:
\[ x \in \left( \frac{3\pi}{4} + 2k\pi, \frac{5\pi}{4} + 2k\pi \right) \quad \text{for any integer } k \]
