Find all the values of x, in between the range [0,2π] for the following equation. 2sin(3x)+3cos(3x)=0

Question

Please solve this math in step by step.I got totally stuck.Can't find the answer..My xm is on 15 th.

Answer 1

2sin(3x)+3cos(3x) = 0

2sin(3x)=-3cos(3x)

sin(3x)/cos(3x) = tan(3x) = -3/2

x = (tan^-1(-3/2) + nπ)/3 = (-tan^-1(3/2) + nπ)/3

To stay within [0,2π], n = 1,2,3,4

Answer 2

2sin(3x) = - 3cos(3x)

(sin(3x))/ (cos (3x)) = -3/2

tan(3x) = -3/2

tan(3x) = tan θ , where θ = tan^-1(-3/2)

3x = κπ + θ, κ any integer

x = κ(π/3) + θ/3

x = κ(π/3) + (1/3) tan^-1(-3/2)

x = κ(π/3) − (1/3) tan^-1(3/2)