$$ + \cos ^ { 2 } x + \sin x \cos x

Question

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

UpStudy AI · Accepted Answer

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:

$$
\text{All } x \text{ in the interval } [90^\circ, 270^\circ].
$$
The inequality $$ \cos^2 x + \sin x \cos x < 2 $$ holds true for all $$ x $$ in the interval $$ 90^\circ \leq x \leq 270^\circ $$.

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

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