When solving a simple trigonometric equation, first isolate the trigonometric function, and then determine the values for the _________ itself...? *adding another in progress*
I guess they are looking for variable... that is how I would solve something like tan(theta)+1=0 isolate the trig function tan(theta)=-1 then find theta such that the equation is true...
Okay - internet is being almighty lag-monster today, sorry -
only one of those make sense
what is your guess from the choices
If I asked you to evaluate tan(45 deg) what is 45 deg what does degrees measure ?
"angles" is wrong
oh that was the only option that made sense :p
submitted that before and was told it was incorrect, so I put this up here since I wouldn't know any other answer besides y-coordinate and trig function (both of which were marked wrong apparently)
ok well you said there was another question?
and this was the exact thing that was on your thingy right: "When solving a simple trigonometric equation, first isolate the trigonometric function, and then determine the values for the _________ itself...? "
sorry OS literally lag-crashed
There is but my internet's being laggy and OS is worse haha
well isolate the trig function first... \[\sin(\theta)=\frac{\sqrt{2}}{2}\] then whip out your unit circle and find when sin(theta) is sqrt(2)/2 also sin(theta) is the y-coordinate on the unit circle
so you are scanning the unit circle for the pairs (x,y) where y is sqrt(2)/2
and giving me the angle that corresponds to that y
ALMIGHTY LAG what gimme a sec to read that lol
Oh, it can work like that?
0-0
are you having trouble?
there is two values in (0,360) one is in the first quadrant the other is in the 2nd quadrant such that y is sqrt(2)/2
45 degrees and umm 180+45
180-45*
45 deg is right
135?
and 180-45 or 135 is right
Yay :]
so sin(theta)=sqrt(2)/2 when theta=45 or 135 deg
this is part b right
then you just add +360*k on to both of those and you have all your solutions
oh okay
since sin has period 360 deg
???
I'm going to try this but I don't know if it will work http://prntscr.com/98804t
you know it says missing deg symbol right? and also you didn't put the other solution to the first one
?
oh 135?
remember we got 45 and 135 for the second one and all it is saying is that it wants you to put the deg symbol for that and then for the first one I asked you to put +360k on to both so you should have 45+360k and 135+360k and yes since it is complaining about a deg symbol you will need to put one on 45 and 135 and 360
okay lol
or you can put it around (45+360k) and around (135+360k)
it just took off a point for the first question I posted what the heck.
either my internet lagged or the site is stupid
isolate the trig function
and do exactly what we did for the other one
yes but sin(theta)-3 I don't know how to solve
OH WAIT \[\frac{\sqrt{3}}{2}\]
wow that was dumb of me
do you know how to solve \[2y-\sqrt{3}=0 \text{ for } y?\] yes add sqrt(3) on both sides then divide both sides by 2
\[\sin(\theta)=\frac{\sqrt{3}}{2}\]
now again y is the sine value
30+360k, 150+360k?
those are both incorrect
when theta is 30 y is 1/2 when theta is 150 y is 1/2
when theta is ____ y is sqrt(3)/2 and when theta is ____ y is sqrt(3)/2 y is the second number in the ordered pair
ugh I messed up pi/3 with ppi/6 sorry
pi*
so you mean 60 and 120?
yes
that would be correct
and please we learned from the last question it wants you to put in those degree symbols
lol ik
anyways peace and good luck
sleepy time
alright thank you and good night :]
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