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Mathematics 15 Online
OpenStudy (anonymous):

Solve trig equation by squaring each side of the equation...can you check if im doing it right so far? http://bamboodock.wacom.com/doodler/7360631c-e5b2-46b0-8a3d-ab4f5986eb8f

OpenStudy (ash2326):

yeah, you are doing right:)

OpenStudy (anonymous):

okay!

OpenStudy (ash2326):

have you solved it further?

OpenStudy (anonymous):

almost !!

OpenStudy (ash2326):

great:D

OpenStudy (anonymous):

is this right>

OpenStudy (anonymous):

is the steps correct?

OpenStudy (ash2326):

@lsugano what's cos 90 or cos \(\pi/2\)?

OpenStudy (anonymous):

0

OpenStudy (anonymous):

cos(x)=0 or 1-cos(x)=0 x= , ,.. or cos(x)=1 x=...,...,.. or x= ..,..,,, can you find thoses angles ?

OpenStudy (ash2326):

so the equation \[\cos x=0\] the solution is \(x=\pi/2\) but there is one more equation \[1-\cos x=0\] what;s the solution of this one?

OpenStudy (anonymous):

what...why is there 1-cosx = 0? where did that come from

OpenStudy (anonymous):

(a)(b)=0 is equivalent to a =0 or b=0

OpenStudy (ash2326):

if you have an equation \[(x+1)(x+2)=0\] so we have either \[x+1=0\] or \[x+2=0\] so both x=-1 and x=-2 are the solutions the same technique is applied here \[2 \cos x(\cos x-1)=0\] either \(\cos x=0 \) or \(\cos x-1=0\)

OpenStudy (ash2326):

do you get this?

OpenStudy (anonymous):

YES wait i will draw it out so you can check it !

OpenStudy (ash2326):

cool:)

OpenStudy (anonymous):

OpenStudy (ash2326):

yeah, it's correct. Now solve it further

OpenStudy (anonymous):

so i got here

OpenStudy (anonymous):

go on ! You're "looking for" x

OpenStudy (anonymous):

like this!???

OpenStudy (ash2326):

yeah correct, what's the solution for \[\cos x=1\]

OpenStudy (anonymous):

i looked at the unit circle (1,0) cosx=1 so x=0 ?

OpenStudy (ash2326):

yeah, you're correct:)

OpenStudy (anonymous):

and 2 pi

OpenStudy (ash2326):

yes 2\(\pi\) as well

OpenStudy (anonymous):

oh but its o≤x<2pi we stilll include 2pi?

OpenStudy (anonymous):

sorry ! I didn't see that !

OpenStudy (anonymous):

you haven't finished yet !

OpenStudy (anonymous):

you need to check if those values are really solutions ! plug them in the original equation !

OpenStudy (anonymous):

like this...?

OpenStudy (anonymous):

yes ! for all those values !(0,pi/2,3pi/2) and you didn't ask me why you should check !

OpenStudy (anonymous):

we check because not all are solutions of the equation and 3pi/2 is -1 so its not a solution!!!

OpenStudy (anonymous):

But ! We don't do that,always ! this equation ! you could do it differently ! remember that questions about cos(x)+sin(x)=ksin(x+alpha) it can be useful !

OpenStudy (anonymous):

okay!!

OpenStudy (anonymous):

if you solved it that way ! You don't have to check !

OpenStudy (anonymous):

thank you!! saved me!

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