If `2sin^(2)x-3=3 cosx and 90^(@) lt x lt 270^(@)`, the number of values that satisfy the equation is

Options

A. 0
B. 1
C. 2
D. 3

Explanation

To solve the equation \(2\sin^2 x - 3 = 3\cos x\) for \(90^\circ < x < 270^\circ\), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ 2\sin^2 x - 3 = 3\cos x \]

### Step 2: Use the Pythagorean identity
We know that \(\sin^2 x + \cos^2 x = 1\). Therefore, we can express \(\sin^2 x\) in terms of \(\cos x\):
\[ \sin^2 x = 1 - \cos^2 x \]
Substituting this into the equation gives:
\[ 2(1 - \cos^2 x) - 3 = 3\cos x \]

### Step 3: Simplify the equation
Now, distribute and simplify:
\[ 2 - 2\cos^2 x - 3 = 3\cos x \]
\[ -2\cos^2 x - 1 = 3\cos x \]
Rearranging the equation:
\[ 2\cos^2 x + 3\cos x + 1 = 0 \]

### Step 4: Substitute \(y = \cos x\)
Let \(y = \cos x\). The equation becomes:
\[ 2y^2 + 3y + 1 = 0 \]

### Step 5: Factor the quadratic equation
We can factor the quadratic equation:
\[ (2y + 1)(y + 1) = 0 \]
Setting each factor to zero gives:
1. \(2y + 1 = 0 \Rightarrow y = -\frac{1}{2}\)
2. \(y + 1 = 0 \Rightarrow y = -1\)

### Step 6: Find \(x\) values
Now we will find \(x\) for each value of \(y\):
1. For \(y = -\frac{1}{2}\):
\[ \cos x = -\frac{1}{2} \Rightarrow x = 120^\circ \text{ or } 240^\circ \]
2. For \(y = -1\):
\[ \cos x = -1 \Rightarrow x = 180^\circ \]

### Step 7: List all possible values of \(x\)
The values of \(x\) that satisfy the equation within the given range \(90^\circ < x < 270^\circ\) are:
- \(120^\circ\)
- \(180^\circ\)
- \(240^\circ\)

### Step 8: Count the number of solutions
Thus, the total number of values of \(x\) that satisfy the equation is:
\[ \text{Total values} = 3 \]

### Final Answer
The number of values that satisfy the equation is **3**.

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