Need pre-calculus help to solve an equation

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2sin^2 3x-1=0

May 27th, 2015

Let sin (3x) = t

we get:

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

May 27th, 2015

I don't understand how the last step gives us the answer given.

May 27th, 2015

3x = pi/4+ pi*k

x = (pi/4+ pi*k)/3 = pi/12 + (pi*k)/3

May 27th, 2015

Can I add 2pi*k? Or does it have to be pi*k and if so why?

May 27th, 2015

Sorry, I mean 2pi*k. You should add 2pi*k.

May 27th, 2015

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.

May 27th, 2015

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

May 27th, 2015

Okay but how do you figure out the period in this situation?

May 27th, 2015

Because x will be equal to 3pi/4, 5pi/4 and 7pi/4. So, the period will be pi*k/2

May 27th, 2015

Ohh okay. Thank you!

May 27th, 2015

You are welcome

May 27th, 2015

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May 27th, 2015
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