Let sin (3x) = t
2t^2 - 1 = 0
2t^2 = 1
t^2 = 1/2
t = sqrt(1/2) = 1/sqrt(2)
sin (3x) = 1/sqrt(2) and sin(3x) = -1/sqrt(2)
x = pi/12 + (pi k)/3
I don't understand how the last step gives us the answer given.
3x = pi/4+ pi*k
x = (pi/4+ pi*k)/3 = pi/12 + (pi*k)/3
Can I add 2pi*k? Or does it have to be pi*k and if so why?
Sorry, I mean 2pi*k. You should add 2pi*k.
I added 2pi*k and I get pi/12 + 2pi*k/3 but when I checked the answer it says, pi/12 + pi*k/6 and I don't understand why.
Ah, I understand what is happening. The period must be (pi*k)/2.
And the answer will be x = (pi/4+ (pi*k)/2)/3 = pi/12 + (pi*k)/6
Okay but how do you figure out the period in this situation?
Because x will be equal to 3pi/4, 5pi/4 and 7pi/4. So, the period will be pi*k/2
Ohh okay. Thank you!
You are welcome
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