Solve the inequality . ` sin x cos x +(1)/(2) tan x ge 1`

Answer

To solve the inequality \( \sin x \cos x + \frac{1}{2} \tan x \geq 1 \), we will follow these steps:

### Step 1: Rewrite the inequality
We start with the original inequality:
\[ \sin x \cos x + \frac{1}{2} \tan x \geq 1 \]

### Step 2: Use the identity for sine
Recall that \( \sin 2x = 2 \sin x \cos x \). We can rewrite \( \sin x \cos x \) as:
\[ \sin x \cos x = \frac{1}{2} \sin 2x \]
Thus, the inequality becomes:
\[ \frac{1}{2} \sin 2x + \frac{1}{2} \tan x \geq 1 \]

### Step 3: Multiply through by 2
To eliminate the fraction, multiply the entire inequality by 2:
\[ \sin 2x + \tan x \geq 2 \]

### Step 4: Express tangent in terms of sine and cosine
Recall that \( \tan x = \frac{\sin x}{\cos x} \). Substitute this into the inequality:
\[ \sin 2x + \frac{\sin x}{\cos x} \geq 2 \]

### Step 5: Rewrite sine double angle
Using the identity \( \sin 2x = 2 \sin x \cos x \), we can rewrite the inequality as:
\[ 2 \sin x \cos x + \frac{\sin x}{\cos x} \geq 2 \]

### Step 6: Combine terms
To combine the terms, we can find a common denominator:
\[ \frac{2 \sin x \cos^2 x + \sin x}{\cos x} \geq 2 \]
This simplifies to:
\[ \frac{\sin x (2 \cos^2 x + 1)}{\cos x} \geq 2 \]

### Step 7: Rearranging the inequality
Rearranging gives:
\[ \sin x (2 \cos^2 x + 1) \geq 2 \cos x \]

### Step 8: Analyze the inequality
Now we need to analyze the inequality \( \sin x (2 \cos^2 x + 1) - 2 \cos x \geq 0 \).

### Step 9: Finding critical points
To find the values of \( x \) that satisfy the inequality, we can set the left-hand side to zero and solve for \( x \):
\[ \sin x (2 \cos^2 x + 1) - 2 \cos x = 0 \]

### Step 10: Solve for \( x \)
This equation can be solved using numerical methods or graphing techniques, as it may not yield simple algebraic solutions.

### Final Step: Determine intervals
Once we find the critical points, we can test intervals around these points to see where the inequality holds true.