Question

Solution of sinx + sin5x = sin 3x in (0,ℼ/2) are..?

Answer 1

Given Equation:
The given equation is: \[ \sin(x) + \sin(5x) = \sin(3x) \] We can simplify this equation using trigonometric identities.

Step 1: Use the Sum-to-Product Identity:
Apply the sum-to-product identity for sine functions: \[ \sin(A) + \sin(B) = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \] Using this identity for \( \sin(x) + \sin(5x) \), we get: \[ \sin(x) + \sin(5x) = 2 \sin(3x) \cos(2x) \] So, the equation becomes: \[ 2 \sin(3x) \cos(2x) = \sin(3x) \]

Step 2: Factor the Equation:
We can factor out \( \sin(3x) \) from both sides of the equation: \[ \sin(3x) \left( 2 \cos(2x) - 1 \right) = 0 \] This equation is satisfied when either \( \sin(3x) = 0 \) or \( 2 \cos(2x) - 1 = 0 \).

Step 3: Solve for \( \sin(3x) = 0 \):
For \( \sin(3x) = 0 \), the solutions occur when: \[ 3x = n\pi \quad \text{where} \quad n \in \mathbb{Z} \] This gives: \[ x = \frac{n\pi}{3} \] So, the possible solutions for \( \sin(3x) = 0 \) are \( x = 0, \pi, 2\pi, 3\pi, \dots \)

Step 4: Solve for \( 2 \cos(2x) - 1 = 0 \):
For \( 2 \cos(2x) - 1 = 0 \), solve for \( \cos(2x) \): \[ 2 \cos(2x) = 1 \quad \Rightarrow \quad \cos(2x) = \frac{1}{2} \] The solutions to \( \cos(2x) = \frac{1}{2} \) occur when: \[ 2x = \pm \frac{\pi}{3} + 2k\pi \quad \text{where} \quad k \in \mathbb{Z} \] So, we have: \[ x = \pm \frac{\pi}{6} + k\pi \] Therefore, the solutions for \( 2 \cos(2x) - 1 = 0 \) are \( x = \frac{\pi}{6}, \frac{5\pi}{6}, \dots \)

Final Answer:
Combining the solutions from both cases, the possible solutions to the equation \( \sin(x) + \sin(5x) = \sin(3x) \) are: \[ x = 0, \frac{\pi}{6}, \frac{\pi}{3} \] Hence, the solutions are \( \boxed{x = \frac{\pi}{6}, \frac{\pi}{3}} \).