Author Topic: General Solution of Circular Function Equations (Read 2946 times) Tweet Share

mandy · « **Reply #15 on:** February 21, 2010, 03:19:55 pm »

So what you're saying is .. the number in front of the n for tan graphs can be any number? :S

the.watchman · « **Reply #16 on:** February 21, 2010, 03:21:26 pm »

No, it should be 1; 3 excludes some solutions, right?

TrueTears · « **Reply #17 on:** February 21, 2010, 03:25:28 pm »

Quote from: the.watchman on February 21, 2010, 03:21:26 pm

No, it should be 1; 3 excludes some solutions, right?

correct.

mandy · « **Reply #18 on:** February 21, 2010, 03:27:49 pm »

Alrighttt, thank you!

mandy · « **Reply #19 on:** February 21, 2010, 04:46:28 pm »

How would I do this one:

Find the general solution to $2cos(2x + \frac{\pi}{4}) = \sqrt2$ , and hence find all the solutions for $x$ in the interval $(-2\pi, 2\pi)$ .

the.watchman · « **Reply #20 on:** February 21, 2010, 04:55:43 pm »

$2 \cos {(2x+\frac{\pi}{4})} = \sqrt{2}$ , for $x \in [0,2\pi]$ , $2x+\frac{\pi}{4} \in [\frac{\pi}{4}, 4\pi + \frac{\pi}{4}]$

$\cos {(2x+\frac{\pi}{4})} = \frac{\sqrt{2}}{2}$

$2x+\frac{\pi}{4} = 2n\pi + \frac{\pi}{4}, 2n\pi - \frac{\pi}{4}$

$x = n\pi, n\pi - \frac{\pi}{4}$

Now you can find the solutions in the interval required!

mandy · « **Reply #21 on:** February 21, 2010, 05:23:09 pm »

Omg, how do I find the solutions in the interval required, LOL.

the.watchman · « **Reply #22 on:** February 21, 2010, 05:26:57 pm »

LOL:

$x = n\pi, n\pi - \frac{\pi}{4}$ AND $x \in (-2\pi,2\pi)$

so $x = -\pi, -\pi - \frac{\pi}{4}, 0, 0 - \frac{\pi}{4}, \pi, \pi - \frac{\pi}{4}$

$x = -\frac{5\pi}{4}, -\pi, -\frac{\pi}{4}, 0, \frac{3\pi}{4}, \pi$

mandy · « **Reply #23 on:** February 21, 2010, 05:35:14 pm »

Quote from: the.watchman on February 21, 2010, 05:26:57 pm

LOL:

$x = n\pi, n\pi - \frac{\pi}{4}$ AND $x \in (-2\pi,2\pi)$

so $x = -\pi, -\pi - \frac{\pi}{4}, 0, 0 - \frac{\pi}{4}, \pi, \pi - \frac{\pi}{4}$

$x = -\frac{5\pi}{4}, -\pi, -\frac{\pi}{4}, 0, \frac{3\pi}{4}, \pi$

Wow, amazing. I think I need to karma you like ...another 20 times today, but thats not possible and I'd have to wait too long LOL. Thank you again !

the.watchman · « **Reply #24 on:** February 21, 2010, 05:37:07 pm »

OOPS, add $2\pi - \frac{\pi}{4} = \frac{7\pi}{4}$ to the list of solutions

I forgot...

Dw, I like the practice!
I'm getting sick of doing my transformations hw from school

mandy · « **Reply #25 on:** February 21, 2010, 05:47:00 pm »

Haha, oh thank you.

TrueTears · « **Reply #26 on:** February 21, 2010, 05:47:39 pm »

Quote from: mandy on February 21, 2010, 05:35:14 pm

Quote from: the.watchman on February 21, 2010, 05:26:57 pm
LOL:

$x = n\pi, n\pi - \frac{\pi}{4}$ AND $x \in (-2\pi,2\pi)$

so $x = -\pi, -\pi - \frac{\pi}{4}, 0, 0 - \frac{\pi}{4}, \pi, \pi - \frac{\pi}{4}$

$x = -\frac{5\pi}{4}, -\pi, -\frac{\pi}{4}, 0, \frac{3\pi}{4}, \pi$

Wow, amazing. I think I need to karma you like ...another 20 times today, but thats not possible and I'd have to wait too long LOL. Thank you again !

ok let me do it for u

mandy · « **Reply #27 on:** February 21, 2010, 05:55:22 pm »

Wow, the power of being moderator! Thanks TT.

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Author Topic: General Solution of Circular Function Equations (Read 2946 times) Tweet Share

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