Find angles of sin(3x) = 0.5

This question is from Calculus II, polar curves etc.

Please explain how to find all the angle of sin(3x) = 0.5, I can find the first one, which is pi/18.

The question and solution is in the following link (pdf), question 5, I don't know how to get the rest of the angles. I also know how to solve the question, but it's the angles theta that's driving me crazy.

http://equaser.com/download/download.php…

OR

http://www.wolframalpha.com/input/?i=pol…

I appreciate any helpful input.

sin(A) = 1/2 has two solutions in [0, 2pi]. In the first quadrant, there's pi/6, and in the second quadrant pi - pi/6 = 5pi/6.

But there are other solutions too; adding any multiple of 2pi gives another solution.

So in general, A = pi/6 + k*2pi, and A = 5pi/6 + k*2pi, for k any integer, is the solution to sin(A)=1/2.

In your problem, you want sin(3x) = 1/2. Replace A with 3x, and divide by 3.

You probably have established that 3x = pi/6

But, the general angles are:

3x = pi/6 ± 2npi => x = pi/18 ± 2npi/3

or 3x = (pi - pi/6) ± 2npi = 5pi/6 ± 2npi => x = 5pi/18 ± 2npi/3......where n = 1, 2, 3,......

so, x = pi/18, 5pi/18, 13pi/18, 17pi/18, 25pi/18 and 29pi/18.....for 0 ≤ x ≤ 2pi

:)>

sin(3x)=sin(1/6*pi)
3x=1/6* pi +k*2pi or 3x = pi-1/6*pi +k*2pi
x=1/18pi+k*2/3pi or x=5/18pi+k*2/3pi