I have a test tomrow and i cant do some of the problems on the topic please help. i want to know how to do these before my test tomrow

express sin 320degrees as a sin function of a positive acute angle

find the value of cos^2 x - 2cos x= 0

thx

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- Dec 10th 2008, 02:39 PMmadfreshxHelp Me with trigs please
I have a test tomrow and i cant do some of the problems on the topic please help. i want to know how to do these before my test tomrow

express sin 320degrees as a sin function of a positive acute angle

find the value of cos^2 x - 2cos x= 0

thx - Dec 10th 2008, 04:14 PMskeeter
$\displaystyle \sin(320) = -\sin(40)$

320 degrees is 40 degrees below the (+) x-axis

$\displaystyle \cos^2{x} - 2\cos{x}= 0$

factor out $\displaystyle \cos{x}$ ...

$\displaystyle \cos{x}(\cos{x} - 2) = 0$

$\displaystyle \cos{x} = 0$ at $\displaystyle x = \frac{\pi}{2}$ and $\displaystyle x = \frac{3\pi}{2}$

$\displaystyle \cos{x} = 2$ has no solution. - Dec 10th 2008, 05:08 PMmadfreshx
thank you very much

can you also help me with this

sinx = - 1/2 solve for x in neareast degree of 0 equal or greater than x equal or greater 360

also for the second question you asnwered for me i made a mistake

it is deta not x - Dec 10th 2008, 05:29 PMskeeter