Question

Find all solutions to the equation sin(3x)cosx-sinx cos(3x)=0 on the interval [0,2pi]

a- x=0,pi/4,pi/2,3pi/4,pi,3pi/2,2pi
b- x=0,pi/2,pi,3pi/2,2pi
c- x=0,pi,2pi
d- x=0,pi/2,3pi/2

Answer 1

Cos(x)Cos(3x) + sin(x)sin(3x)=0 [0,2pi]

Answer: x= pi/4, 3pi/4, 5pi/4, 7pi/4

explanation:For those who have a different answer choice along with the question.

Answer 2

Please have in mind that sin(A - B) = sin(A)cos(B) - cos(A)sin(B)
So what we do is:
sin(3x)cos(x) - sin(x)cos(3x) = 0 => sin(3x - x) = 0

sin(2x) = 0

2x = 0, π, 2π, 3π, 4π

x = 0, π/2, π, 3π/2, 2π
With this, even though not giving you all the answers, will help you greatly to get them. I know you will get the answers with no problems.