\( + \cos ^ { 2 } x + \sin x \cos x < 2,90 ^ { \circ } \leq x \leq 270 ^ { \circ } \)

It seems that the inequality \( \cos^2 x + \sin x \cos x < 2 \) cannot be solved directly using the tools available. However, I can guide you through the steps to analyze and solve this inequality manually. ### Step 1: Understand the Range of the Functions The functions \( \sin x \) and \( \cos x \) oscillate between -1 and 1. Therefore, \( \cos^2 x \) will range from 0 to 1, and \( \sin x \cos x \) will range from -0.5 to 0.5. ### Step 2: Analyze the Left Side of the Inequality The left side of the inequality is \( \cos^2 x + \sin x \cos x \). 1. **Maximum of \( \cos^2 x \)**: The maximum value is 1 when \( x = 0^\circ, 180^\circ, 360^\circ \). 2. **Maximum of \( \sin x \cos x \)**: This can be rewritten using the double angle identity: \[ \sin x \cos x = \frac{1}{2} \sin(2x) \] The maximum value of \( \frac{1}{2} \sin(2x) \) is \( \frac{1}{2} \) when \( 2x = 90^\circ + k \cdot 360^\circ \) for integers \( k \). ### Step 3: Combine the Maximum Values The maximum value of \( \cos^2 x + \sin x \cos x \) is: \[ 1 + \frac{1}{2} = 1.5 \] This is less than 2, so the inequality \( \cos^2 x + \sin x \cos x < 2 \) holds for all \( x \). ### Step 4: Determine the Interval The given interval is \( 90^\circ \leq x \leq 270^\circ \). ### Conclusion Since the inequality \( \cos^2 x + \sin x \cos x < 2 \) holds for all \( x \), it also holds for the specified interval. Therefore, the solution to the inequality is: \[ \text{All } x \text{ in the interval } [90^\circ, 270^\circ]. \]