If `2cos^(2)x+3sinx-3=0,0lexle180^(@)`, then x =

Options

A. `30^(@),90^(@),150^(@)`
B. `60^(@),120^(@),180^(@)`
C. `0^(@),30^(@),150^(@)`
D. `45^(@),90^(@),135^(@)`

Answer

To solve the equation \(2\cos^2 x + 3\sin x - 3 = 0\) for \(0 \leq x \leq 180^\circ\), we can follow these steps:

### Step 1: Rewrite the equation using the Pythagorean identity
We know that \(\cos^2 x = 1 - \sin^2 x\). We can substitute this into the equation:
\[ 2(1 - \sin^2 x) + 3\sin x - 3 = 0 \]

### Step 2: Simplify the equation
Distributing the \(2\) gives:
\[ 2 - 2\sin^2 x + 3\sin x - 3 = 0 \]
Combining like terms leads to:
\[ -2\sin^2 x + 3\sin x - 1 = 0 \]
Multiplying through by \(-1\) to make the leading coefficient positive:
\[ 2\sin^2 x - 3\sin x + 1 = 0 \]

### Step 3: Solve the quadratic equation
We can use the quadratic formula \( \sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \(a = 2\), \(b = -3\), and \(c = 1\):
\[ \sin x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} \]
Calculating the discriminant:
\[ \sin x = \frac{3 \pm \sqrt{9 - 8}}{4} \]
This simplifies to:
\[ \sin x = \frac{3 \pm 1}{4} \]

### Step 4: Find the values of \(\sin x\)
Calculating the two possible values:
1. \( \sin x = \frac{4}{4} = 1 \)
2. \( \sin x = \frac{2}{4} = \frac{1}{2} \)

### Step 5: Determine the angles corresponding to these sine values
1. For \( \sin x = 1 \):
- The angle \(x = 90^\circ\).

2. For \( \sin x = \frac{1}{2} \):
- The angles are \(x = 30^\circ\) and \(x = 150^\circ\) (since sine is positive in both the first and second quadrants).

### Step 6: Compile the solutions
Thus, the solutions for \(x\) in the interval \(0 \leq x \leq 180^\circ\) are:
\[ x = 30^\circ, 90^\circ, 150^\circ \]

### Final Answer
The values of \(x\) are \(30^\circ\), \(90^\circ\), and \(150^\circ\).

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