Hi, need help understanding 2 problems.

Give all the solutions to the following problems, answer in radians.

$\displaystyle \sin(3x)=\sqrt{3}/2$

and

$\displaystyle \cos(x-pi/4)=1/\sqrt{2}$

Thanks.

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- May 29th 2008, 07:23 AMSmifsretRadians
Hi, need help understanding 2 problems.

Give all the solutions to the following problems, answer in radians.

$\displaystyle \sin(3x)=\sqrt{3}/2$

and

$\displaystyle \cos(x-pi/4)=1/\sqrt{2}$

Thanks. - May 29th 2008, 07:34 AMcolby2152
- May 29th 2008, 08:27 AMSmifsret
Thanks, is the second one pi / 4 then?

Does pi/4 count as an answer (if its right) or do they want me to write a number? - May 29th 2008, 09:51 AMSmifsret
sin(3x)= pi/3

x1. sin(x) = pi/9 + 2 * pi * n?

x2. sin(x) = pi-pi/9 + 2 * pi * n?

Is that correct? - May 29th 2008, 10:09 AMSmifsret
Ahh the question is to answer EXACTLY in radians and I was told it isn't exactly unless you write it like that.

Also the formula + n * 2 * pi I was told to add because n is the number of times it has gone around the circle.

Hope I explained that decently.

But I understand that I shouldn't have addad sin to the solution, thank you, as always I wasn't aware that it mattered. - May 29th 2008, 10:13 AMSmifsret
For the second part

x1 = pi/2

x2 = -pi/2

Correct? - May 29th 2008, 12:22 PMtopsquark
- May 29th 2008, 12:48 PMSmifsret
You wrote that

$\displaystyle x - \frac{\pi}{4} = \frac{\pi}{6} $

i think that its

$\displaystyle 1/\sqrt{2}=0.70711 = 45 degrees = \frac{\pi}{4}$

Or am I completely wrong?

- May 29th 2008, 12:53 PMtopsquark
- May 29th 2008, 12:56 PMSmifsret
Thank you very much for the help, does that mean that my previous answers were correct?