Author Topic: General Solutions Trigonometry (Read 2073 times) Tweet Share

wildareal · « **on:** May 01, 2010, 09:14:22 pm »

Hi there,

Which way do you guys do General Solutions, for example:

√2sin(2x)+1=0

Thanks

TrueTears · « **Reply #1 on:** May 01, 2010, 09:21:55 pm »

hi there mate,

sin(2x) = -1/rt[2]

then find the 2 basic values for x and then add 2pi/2+2pi(n) to each basic solution where n in Z

stonecold · « **Reply #2 on:** May 01, 2010, 09:26:18 pm »

Use $x=\pi n + (-1)^nsin^-^1(a)$

$sin(2x)=-\frac{\sqrt{2}}{2}$

$2x = \pi n + (-1)^nsin^-^1(-\frac{\sqrt{2}}{2})$

$2x = \pi n + (-1)^n\frac{-\pi}{4}$

$2x = \frac{4\pi n - (-1)^n\pi}{4}$

$2x = \frac{\pi(4n - (-1)^n)}{4}$

$x = \frac{\pi(4n - (-1)^n)}{8}$ where $n\in Z$

vea · « **Reply #3 on:** May 01, 2010, 09:40:33 pm »

$\sqrt{2}sin(2x)+1=0$

$sin(2x)=-\frac{\sqrt{2}}{2}$

$2x=2n\pi+sin^-^1(-\frac{\sqrt{2}}{2}), (2n+1)\pi-sin^-^1(-\frac{\sqrt{2}}{2})$ , where n is an integer

$2x=2n\pi+\frac{-\pi}{4}, (2n+1)\pi-\frac{-\pi}{4}$ , where n is an integer

$x=n\pi-\frac{\pi}{8}, n\pi+\frac{\pi}{2}+\frac{\pi}{8}$ , where n is an integer

$x=n\pi-\frac{\pi}{8}, n\pi+\frac{5\pi}{8}$ , where n is an integer

PS. Don't forget the "where n is an integer" or "where n is an element of z" or else you may get marks deducted.

stonecold · « **Reply #4 on:** May 01, 2010, 10:17:20 pm »

Yeah, the where n=Z is probably important, could cost you a mark.

Also remember the relevant ranges of sin, cos and tan. Don't make the same mistake I did.

Fixed it up now.

$sin^-^1$ has range $[-\frac{\pi}{2},\frac{\pi}{2}]$

$tan^-^1$ has range $(-\frac{\pi}{2},\frac{\pi}{2})$

$cos^-^1$ has range $[0,\pi]$

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Author Topic: General Solutions Trigonometry (Read 2073 times) Tweet Share

wildareal

General Solutions Trigonometry

TrueTears

Re: General Solutions Trigonometry

stonecold

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vea

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stonecold

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