If `Cos20^0=k` and `Cosx=2k^2-1`, then the possible values of x between `0^0 and 360^0` are

Options

A. `140^(@)`
B. `40^(@) " and "140^(@)`
C. `40^(@) " and "320^(@)`
D. `50^(@) " and "130^(@)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Substitute the value of \( k \)
Given that \( \cos 20^\circ = k \), we can substitute this into the equation for \( \cos x \):
\[ \cos x = 2k^2 - 1 \]
Substituting \( k \):
\[ \cos x = 2(\cos 20^\circ)^2 - 1 \]

### Step 2: Use the double angle formula
We can use the double angle formula for cosine, which states that:
\[ \cos 2\theta = 2\cos^2 \theta - 1 \]
In our case, let \( \theta = 20^\circ \):
\[ \cos x = \cos 40^\circ \]

### Step 3: Set up the equation
Now we have:
\[ \cos x = \cos 40^\circ \]
This means that \( x \) can take values where the cosine function is equal to \( \cos 40^\circ \).

### Step 4: Find general solutions for \( x \)
The general solutions for \( \cos x = \cos A \) are given by:
\[ x = 2n\pi \pm A \]
where \( n \) is any integer. For our case:
\[ x = 2n\pi \pm 40^\circ \]

### Step 5: Determine specific values of \( x \) in the range \( 0^\circ \) to \( 360^\circ \)
1. For \( n = 0 \):
- \( x = 40^\circ \)
- \( x = 360^\circ - 40^\circ = 320^\circ \)

2. For \( n = 1 \):
- \( x = 2(180^\circ) + 40^\circ = 400^\circ \) (not in the range)
- \( x = 2(180^\circ) - 40^\circ = 320^\circ \) (already found)

Thus, the possible values of \( x \) between \( 0^\circ \) and \( 360^\circ \) are:
\[ \boxed{40^\circ \text{ and } 320^\circ} \]