solve the equaton :

sin 3x + sin 5x + sin 7x =0 for pi/2

To solve the equation sin(3x) + sin(5x) + sin(7x) = 0 for the interval [0, π/2], we can approach it step by step. This equation involves multiple sine functions, which may seem complex, but we can simplify our process using trigonometric identities and properties.

Understanding the Equation

The equation we have is:

sin(3x) + sin(5x) + sin(7x) = 0

To find the solutions, we can utilize the sine function's properties and the fact that it oscillates between -1 and 1. We will also consider the specific interval of [0, π/2].

Using Trigonometric Identities

One helpful identity for sums of sine functions is:

sin(A) + sin(B) = 2 * sin((A + B)/2) * cos((A - B)/2)

We can use this identity to combine the sine terms in pairs. Let’s first combine sin(3x) and sin(7x):

• Let A = 7x and B = 3x.
• Applying the identity gives us:

  sin(3x) + sin(7x) = 2 * sin(5x) * cos(2x)

Now we rewrite the equation:

2 * sin(5x) * cos(2x) + sin(5x) = 0

Factoring the Expression

Next, we can factor out sin(5x):

sin(5x) * (2 * cos(2x) + 1) = 0

Finding the Roots

This gives us two cases to solve:

• Case 1: sin(5x) = 0
• Case 2: 2 * cos(2x) + 1 = 0

Case 1: Solving sin(5x) = 0

The general solution for sin(θ) = 0 is θ = nπ, where n is an integer. Thus:

5x = nπ ⟹ x = nπ/5

Now, we need to find values of n such that x falls within [0, π/2]:

• For n = 0: x = 0
• For n = 1: x = π/5
• For n = 2: x = 2π/5
• For n = 3: x = 3π/5 (this exceeds π/2)

Case 2: Solving 2 * cos(2x) + 1 = 0

From this, we can simplify to:

cos(2x) = -1/2

The general solutions for cos(θ) = -1/2 occur at:

• θ = (2n + 1)π/3.

So, we set:

2x = (2n + 1)π/3 ⟹ x = (2n + 1)π/6.

Now we check which values of n fit within [0, π/2]:

• For n = 0: x = π/6
• For n = 1: x = π/2 (which is the boundary of our interval)
• For n = 2: x = 5π/6 (exceeds π/2)

Summary of Solutions

Combining the results from both cases, we find the solutions to the equation sin(3x) + sin(5x) + sin(7x) = 0 within the interval [0, π/2] are:

• x = 0
• x = π/5
• x = π/6
• x = π/2

These solutions give us the points where the original equation holds true, illustrating the properties of the sine function and how it can be manipulated using trigonometric identities.