Number of solution `(s)` of the equation `((2-cos^2x)/(sinx))^3+((3-cos2x)/(sinx))=0` is

Options

A. 0
B. 1
C. 2
D. infinite

Explanation

To solve the equation

\[ \left(\frac{2 - \cos^2 x}{\sin x}\right)^3 + \left(\frac{3 - \cos 2x}{\sin x}\right) = 0, \]

we will follow these steps:

### Step 1: Rewrite the equation using trigonometric identities

We know that

\[ \cos 2x = 1 - 2\sin^2 x. \]

Substituting this into the equation gives:

\[ \left(\frac{2 - \cos^2 x}{\sin x}\right)^3 + \left(\frac{3 - (1 - 2\sin^2 x)}{\sin x}\right) = 0. \]

This simplifies to:

\[ \left(\frac{2 - \cos^2 x}{\sin x}\right)^3 + \left(\frac{2 + 2\sin^2 x}{\sin x}\right) = 0. \]

### Step 2: Substitute \(\cos^2 x\)

Using the identity \(\cos^2 x = 1 - \sin^2 x\), we can rewrite the first term:

\[ \left(\frac{2 - (1 - \sin^2 x)}{\sin x}\right)^3 + \left(\frac{2 + 2\sin^2 x}{\sin x}\right) = 0. \]

This simplifies to:

\[ \left(\frac{1 + \sin^2 x}{\sin x}\right)^3 + \left(\frac{2(1 + \sin^2 x)}{\sin x}\right) = 0. \]

### Step 3: Let \(t = \frac{1 + \sin^2 x}{\sin x}\)

Now, we can substitute \(t\) into the equation:

\[ t^3 + 2t = 0. \]

### Step 4: Factor the equation

Factoring out \(t\):

\[ t(t^2 + 2) = 0. \]

### Step 5: Solve for \(t\)

Setting each factor to zero gives us:

1. \(t = 0\)
2. \(t^2 + 2 = 0\) (which has no real solutions since \(t^2 + 2\) is always positive).

### Step 6: Solve \(t = 0\)

Substituting back for \(t\):

\[ \frac{1 + \sin^2 x}{\sin x} = 0. \]

This implies:

\[ 1 + \sin^2 x = 0. \]

Since \(\sin^2 x\) is always non-negative, \(1 + \sin^2 x = 0\) has no solutions.

### Conclusion

Since there are no valid solutions for \(t\), we conclude that the original equation has no solutions.

Thus, the number of solutions \(s\) of the equation is:

\[ \boxed{0}. \]