Let `y = cosx (cosx - cos 3x)`. Then y is

Options

A. `ge 0` only when `x ge 0`
B. `le 0`, for all real x
C. `ge 0` for all real x
D. `le 0` only when `x le 0`

Answer

To solve the problem, we need to analyze the expression given for \( y \):

\[ y = \cos x (\cos x - \cos 3x) \]

### Step 1: Substitute the formula for \( \cos 3x \)

We know the trigonometric identity for \( \cos 3x \):
\[ \cos 3x = 4 \cos^3 x - 3 \cos x \]

Now, substitute this into the expression for \( y \):

\[ y = \cos x \left( \cos x - (4 \cos^3 x - 3 \cos x) \right) \]

### Step 2: Simplify the expression

Now simplify the expression inside the parentheses:

\[ y = \cos x \left( \cos x - 4 \cos^3 x + 3 \cos x \right) \]
\[ y = \cos x \left( 4 \cos x - 4 \cos^3 x \right) \]

### Step 3: Factor out common terms

We can factor out \( 4 \cos x \):

\[ y = \cos x \cdot 4 \cos x (1 - \cos^2 x) \]
\[ y = 4 \cos^2 x (1 - \cos^2 x) \]

### Step 4: Use the Pythagorean identity

Using the identity \( 1 - \cos^2 x = \sin^2 x \):

\[ y = 4 \cos^2 x \sin^2 x \]

### Step 5: Analyze the nature of \( y \)

Since both \( \cos^2 x \) and \( \sin^2 x \) are always non-negative (i.e., they are squares), their product \( 4 \cos^2 x \sin^2 x \) is also non-negative. The only time \( y \) can be zero is when either \( \cos x = 0 \) or \( \sin x = 0 \).

Thus, we conclude that:

- \( y \) is always non-negative.
- \( y \) can be zero, but it cannot be negative.

### Final Answer

Therefore, the nature of \( y \) is that it is always non-negative.