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Let's break down the problem step-by-step.

The given equation for \( x \) is:

\[
x = \frac{\sin(75^\circ) \cos(75^\circ)}{3 + \cos(18000^\circ)}
\]

### Step 1: Simplifying \(\cos(18000^\circ)\)

The cosine function has a period of \(360^\circ\), so we can reduce the angle modulo \(360^\circ\):

\[
18000^\circ \div 360^\circ = 50
\]

Thus,

\[
18000^\circ \equiv 0^\circ \pmod{360^\circ}
\]

Therefore,

\[
\cos(18000^\circ) = \cos(0^\circ) = 1
\]

### Step 2: Updating the expression for \( x \)

Now the expression for \( x \) becomes:

\[
x = \frac{\sin(75^\circ) \cos(75^\circ)}{3 + 1} = \frac{\sin(75^\circ) \cos(75^\circ)}{4}
\]

### Step 3: Using a trigonometric identity

We can use the identity:

\[
\sin(2\theta) = 2 \sin(\theta) \cos(\theta)
\]

For \(\theta = 75^\circ\), this gives:

\[
\sin(150^\circ) = 2 \sin(75^\circ) \cos(75^\circ)
\]

Thus, the expression for \(x\) becomes:

\[
x = \frac{\sin(150^\circ)}{8}
\]

Since \(\sin(150^\circ) = \sin(30^\circ) = \frac{1}{2}\), we get:

\[
x = \frac{1}{2 \times 8} = \frac{1}{16}
\]

### Step 4: Finding \(\log_2 x\)

We need to compute:

\[
\log_2 \left(\frac{1}{16}\right)
\]

Since \(\frac{1}{16} = 2^{-4}\), we have:

\[
\log_2 \left(\frac{1}{16}\right) = \log_2 (2^{-4}) = -4
\]

Thus, the final result is:

\[
\log_2 x = -4
\]

Would you like more details on any step or have any further questions?

### Related Questions:
1. How do you simplify trigonometric expressions involving large angles?
2. What is the period of cosine and sine functions?
3. How does the identity \(\sin(2\theta) = 2 \sin(\theta) \cos(\theta)\) work?
4. Can you explain logarithmic properties with base 2 in more detail?
5. What other trigonometric identities can be useful in solving complex trigonometric expressions?

**Tip**: Always reduce large angles in trigonometric functions by their period, \(360^\circ\), to simplify calculations.

Find log2(x) where x = (sin(75°) cos(75°)) / (3 + cos(18000°))

Learn how to compute log2(x) for a trigonometric expression that involves angle reduction. This problem uses key trigonometric identities and logarithmic properties to simplify and find the solution. Suitable for students in Grade 11-12.

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