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what do you need help with?
i will attach the file, give me a second
hmm, try replacing cos^2 with 1 - sin^2
which attachment are you on
the first one...
\[3\sin(\theta)+3 = 2\cos^2(\theta)\]
i did that
ok, it looks like you can factor that now if you put it in ax^2+bx+c form , with the x being sin
right, from the unit circle , there is the property, sin^2 + cos^2 = 1, that is what we did
then just simplify that down to
ok got it
so you can factor that like usual quadratics , let v=sin(theta) 2v^2 + 3v + 1 = 0
ok give me a sec
replace v back with sin(theta) \[(\sin \theta + 1)(2\sin \theta +1) = 0\]
good so far?
that is true if either of those parenthesis terms are zero, so you have \[\sin(\theta) = -1~~~or~~~\sin(\theta) = -1/2\]
do you know how to figure those from the unit circle?
yeah, but im kind of weak in my unit circle so i might get it wrong
Ill tell ya, soon as you get that picture, everything clicks good...
ok should I gust google unit circle
do you want a quick rundown of it, how to determine angles
no i got it
i just needed a brush up
the picture is fine
aight, what are the answers for this one
the answers are -pi/2, 3pi/2, -pi/6, 7p/6
read the problem one more time through...
oh so it is just 3pi/2 and 7pi/6
not quite... lets look at the first one... sin(theta) = -1 Each point on the circle is the coordinate (cos , sin) , the y coordinate is the sin of the angle... y is -1 at 270 or 3pi/2
for sin(theta) = -1/2 there are 2 angle values where the y coordinate is -1/2
i remember taking trig and never got how angles were figured till this circle pic , after the class was over
radius is 1, so sin = opp/hyp = opp/1 = opp , or the y coordinate
ok that makes sense
thats helpful thanks
so the solutions to the equation are?