The number of solutions of the equation ` 4 sin^(2) x + tan^(2)x + cot^(2)x + "cosec"^(2)x = 6 ` in `[0, 2 pi]`

Options

A. ` n pi pm (pi)/(4) `
B. ` 2n pi pm (pi)/(4) `
C. ` n pi + (pi)/(3) `
D. ` n pi - (pi)/(6) `

Answer

To solve the equation \( 4 \sin^2 x + \tan^2 x + \cot^2 x + \csc^2 x = 6 \) in the interval \([0, 2\pi]\), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ 4 \sin^2 x + \tan^2 x + \cot^2 x + \csc^2 x = 6 \]
We know that:
\[ \tan^2 x = \frac{\sin^2 x}{\cos^2 x}, \quad \cot^2 x = \frac{\cos^2 x}{\sin^2 x}, \quad \text{and} \quad \csc^2 x = \frac{1}{\sin^2 x} \]
Substituting these into the equation gives:
\[ 4 \sin^2 x + \frac{\sin^2 x}{\cos^2 x} + \frac{\cos^2 x}{\sin^2 x} + \frac{1}{\sin^2 x} = 6 \]

### Step 2: Simplify the equation
To simplify, we can multiply through by \(\sin^2 x \cos^2 x\) (assuming \(\sin x \neq 0\) and \(\cos x \neq 0\)):
\[ 4 \sin^4 x \cos^2 x + \sin^4 x + \cos^4 x + \cos^2 x = 6 \sin^2 x \cos^2 x \]

### Step 3: Use identities
We can use the identity \(\sin^2 x + \cos^2 x = 1\) to rewrite \(\sin^4 x + \cos^4 x\):
\[ \sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - 2\sin^2 x \cos^2 x \]
Substituting this back into our equation gives:
\[ 4 \sin^4 x \cos^2 x + (1 - 2\sin^2 x \cos^2 x) + \cos^2 x = 6 \sin^2 x \cos^2 x \]

### Step 4: Rearranging
Rearranging the equation:
\[ 4 \sin^4 x \cos^2 x + 1 - 2\sin^2 x \cos^2 x + \cos^2 x - 6 \sin^2 x \cos^2 x = 0 \]
This simplifies to:
\[ 4 \sin^4 x \cos^2 x + 1 - 8 \sin^2 x \cos^2 x = 0 \]

### Step 5: Solve for \(\sin^2 x\)
Let \(y = \sin^2 x\), then \(\cos^2 x = 1 - y\):
\[ 4y^2(1 - y) + 1 - 8y(1 - y) = 0 \]
Expanding this gives:
\[ 4y^2 - 4y^3 + 1 - 8y + 8y^2 = 0 \]
Combining like terms:
\[ -4y^3 + 12y^2 - 8y + 1 = 0 \]

### Step 6: Find roots
We can use numerical methods or graphing to find the roots of this cubic equation. Alternatively, we can use the Rational Root Theorem to find possible rational roots.

### Step 7: Count solutions
After finding the roots of the cubic equation, we determine the corresponding values of \(x\) in the interval \([0, 2\pi]\).

### Step 8: Final count
Each valid \(y\) (where \(0 \leq y \leq 1\)) corresponds to two values of \(x\) in \([0, 2\pi]\) due to the periodic nature of sine and cosine.