If `0lexltpi/2,` then the number of values of `x` for which `sinx-sin2x+sin3x=0` is

Options

A. 2
B. 1
C. 3
D. 4

Answer

To solve the equation \( \sin x - \sin 2x + \sin 3x = 0 \) in the interval \( 0 < x < \frac{\pi}{2} \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \sin x - \sin 2x + \sin 3x = 0 \]
We can rearrange it to:
\[ \sin x + \sin 3x = \sin 2x \]

### Step 2: Use the sine addition formula
Using the sine addition formula, we can express \( \sin 3x \) as:
\[ \sin 3x = 3 \sin x - 4 \sin^3 x \]
Now, substituting this into our equation gives:
\[ \sin x + (3 \sin x - 4 \sin^3 x) = \sin 2x \]
We also know that \( \sin 2x = 2 \sin x \cos x \). Thus, we can rewrite the equation as:
\[ 4 \sin x - 4 \sin^3 x = 2 \sin x \cos x \]

### Step 3: Factor out common terms
Factoring out \( 2 \sin x \) from both sides, we have:
\[ 2 \sin x (2 - 2 \sin^2 x) = 2 \sin x \cos x \]
Dividing both sides by \( 2 \sin x \) (noting that \( \sin x \neq 0 \) in the interval \( 0 < x < \frac{\pi}{2} \)):
\[ 2 - 2 \sin^2 x = \cos x \]

### Step 4: Use the Pythagorean identity
Using the identity \( \cos x = \sqrt{1 - \sin^2 x} \), we can substitute into the equation:
\[ 2 - 2 \sin^2 x = \sqrt{1 - \sin^2 x} \]
Let \( y = \sin x \). The equation becomes:
\[ 2 - 2y^2 = \sqrt{1 - y^2} \]

### Step 5: Square both sides
Squaring both sides to eliminate the square root gives:
\[ (2 - 2y^2)^2 = 1 - y^2 \]
Expanding the left side:
\[ 4(1 - y^2)^2 = 1 - y^2 \]
This simplifies to:
\[ 4(4y^4 - 4y^2 + 1) = 1 - y^2 \]
Expanding and rearranging gives:
\[ 16y^4 - 16y^2 + 4 - 1 + y^2 = 0 \]
\[ 16y^4 - 15y^2 + 3 = 0 \]

### Step 6: Let \( z = y^2 \)
Let \( z = y^2 \), then we have:
\[ 16z^2 - 15z + 3 = 0 \]
Using the quadratic formula \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ z = \frac{15 \pm \sqrt{(-15)^2 - 4 \cdot 16 \cdot 3}}{2 \cdot 16} \]
Calculating the discriminant:
\[ = \frac{15 \pm \sqrt{225 - 192}}{32} \]
\[ = \frac{15 \pm \sqrt{33}}{32} \]

### Step 7: Find values of \( y \)
Since \( y = \sin x \), we need to find \( y \) values:
\[ y^2 = \frac{15 \pm \sqrt{33}}{32} \]
We need to check which of these values fall within the range \( 0 < y < 1 \).

### Step 8: Count valid solutions
We find the number of valid \( y \) values that correspond to \( x \) values in the interval \( 0 < x < \frac{\pi}{2} \). Each valid \( y \) will correspond to a unique \( x \) in this range.

### Final Answer
After evaluating the valid roots, we find that there are **two values** of \( x \) that satisfy the original equation in the given interval.