Question

For x ∈ (0, π), the equation sin x + 2sin2x −sin3x = 3 has 

• A. Infinitely many solutions
• B. Three solutions
• C. One solution
• D. No solution

Answer 1

The Correct Option is D

To determine the number of solutions for the equation \( \sin x + 2\sin^2x - \sin^3x = 3 \) in the interval \( x \in (0, \pi) \), we need to examine the behavior of the given trigonometric expression.

Let's analyze the equation step by step:

1. First, consider the range of each component in the equation:
   • \(\sin x\) ranges from 0 to 1 for \( x \in (0, \pi) \).
   • \(\sin^2x\) similarly ranges from 0 to 1.
   • \(\sin^3x\) again ranges from 0 to 1 as \( \sin x \) ranges from 0 to 1.
2. Now, consider the maximum value the entire expression \( \sin x + 2\sin^2x - \sin^3x \) can take:
   • At its extremes, \(\sin x = 1\), so \( \sin x + 2\sin^2x - \sin^3x = 1 + 2(1)^2 - (1)^3 = 1 + 2 - 1 = 2 \).
   • The minimum value can occur when \(\sin x = 0\), giving a result of \( 0 + 2 \times 0^2 - 0^3 = 0 \).
3. Thus, the expression \( \sin x + 2\sin^2x - \sin^3x \) ranges from 0 to 2.
4. The value 3 is completely outside of this range (0 to 2). Therefore, it is impossible for the expression to equal 3.

Consequently, the equation \( \sin x + 2\sin^2x - \sin^3x = 3 \) has no solution for \( x \in (0, \pi) \).