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
yeah, you are doing right:)
okay!
have you solved it further?
almost !!
great:D
is this right>
is the steps correct?
@lsugano what's cos 90 or cos \(\pi/2\)?
0
cos(x)=0 or 1-cos(x)=0 x= , ,.. or cos(x)=1 x=...,...,.. or x= ..,..,,, can you find thoses angles ?
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?
what...why is there 1-cosx = 0? where did that come from
(a)(b)=0 is equivalent to a =0 or b=0
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 \(\cos x-1=0\)
do you get this?
YES wait i will draw it out so you can check it !
cool:)
yeah, it's correct. Now solve it further
so i got here
go on ! You're "looking for" x
like this!???
yeah correct, what's the solution for \[\cos x=1\]
i looked at the unit circle (1,0) cosx=1 so x=0 ?
yeah, you're correct:)
and 2 pi
yes 2\(\pi\) as well
oh but its o≤x<2pi we stilll include 2pi?
sorry ! I didn't see that !
you haven't finished yet !
you need to check if those values are really solutions ! plug them in the original equation !
like this...?
yes ! for all those values !(0,pi/2,3pi/2) and you didn't ask me why you should check !
we check because not all are solutions of the equation and 3pi/2 is -1 so its not a solution!!!
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 !
okay!!
if you solved it that way ! You don't have to check !
thank you!! saved me!
Join our real-time social learning platform and learn together with your friends!