• May 29th 2008, 08:23 AM
Smifsret
Hi, need help understanding 2 problems.

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

and

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

Thanks.
• May 29th 2008, 08:34 AM
colby2152
Quote:

Originally Posted by Smifsret
Hi, need help understanding 2 problems.

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

and

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

Thanks.

You need to know what angle for sine gives the value $\frac{\sqrt{3}]{2}$. This happens to be $\frac{\pi}{3}$

Solve for $3x = \frac{\pi}{3}$
• May 29th 2008, 09:27 AM
Smifsret
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, 10:51 AM
Smifsret
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, 11:09 AM
Smifsret
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, 11:13 AM
Smifsret
For the second part

x1 = pi/2

x2 = -pi/2

Correct?
• May 29th 2008, 01:22 PM
topsquark
Quote:

Originally Posted by Smifsret
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.

I'm sorry, I should have realized that your x1 and x2 were solutions to the first problem, not solutions to problems 1 and 2.

In that case, yes, those solutions are correct (with x replacing your sin(x).)

-Dan
• May 29th 2008, 01:48 PM
Smifsret
You wrote that
$x - \frac{\pi}{4} = \frac{\pi}{6}$
i think that its
$1/\sqrt{2}=0.70711 = 45 degrees = \frac{\pi}{4}$

Or am I completely wrong?
• May 29th 2008, 01:53 PM
topsquark
Quote:

Originally Posted by Smifsret
You wrote that
$x - \frac{\pi}{4} = \frac{\pi}{6}$
i think that its
$1/\sqrt{2}=0.70711 = 45 degrees = \frac{\pi}{4}$

Or am I completely wrong?

(Doh) My apologies again. You are correct.

I'm going back to bed. (Headbang)

-Dan
• May 29th 2008, 01:56 PM
Smifsret
Thank you very much for the help, does that mean that my previous answers were correct?