Question

Solve in the interval -90
`\leq`x
`\leq`90, sin(5x) + sin(-7x) = 0. Explaining your full method.

Answer 1

-75°, -45°, -15°, 0°, 15°, 45°, 75°

Given:

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

Common breakdown:

sin(6x-x) + sin(-6x-x) = 0

sin(6x-x) - sin(6x+x) = 0

Use the formula:

sin(A ± B) = sin(A)cos(B) ± cos(B)sin(A)

Evaluate:

[sin(6x)cos(x) - cos(6x)sin(x)] - [sin(6x)cos(x) + cos(6x)sin(x)]

sin(6x)cos(x) - cos(6x)sin(x) - sin(6x)cos(x) - cos(6x)sin(x)

-2cos(6x)sin(x) = 0

Solve them by parts:

cos(6x) = 0, sin(x) = 0

• For cos(6x)

As -90<x<90 so -540<x<540

6x = -450, -270, -90, 90, 270, 450

x = -75, -45, -15, 15, 45, 75

• For sin(x)

sin(x) = 0

x = 0 only.