anonymous
  • anonymous
I have solved a trig problem and want to see if I got the correct answer.
Mathematics
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anonymous
  • anonymous
I have solved a trig problem and want to see if I got the correct answer.
Mathematics
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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anonymous
  • anonymous
anonymous
  • anonymous
jdoe0001
  • jdoe0001
so..what did you get for 7?

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anonymous
  • anonymous
so each question has multiple solutions
jdoe0001
  • jdoe0001
right
anonymous
  • anonymous
For question #7, the solutions I got were\[\pi/4,2\pi/3 \]
anonymous
  • anonymous
and\[5\pi/12,2\pi/3n\]
jdoe0001
  • jdoe0001
well. you may not need the 3n part, since you're only constrained to \((0,2\pi)\)
anonymous
  • anonymous
ok
jdoe0001
  • jdoe0001
\(\bf 2cos(3\theta)+\sqrt{2}=0\implies 2cos(3\theta)=-\sqrt{2} \\ \quad \\ cos(3\theta)=-\cfrac{\sqrt{2}}{2}\implies 3\theta=cos^{-1}\left( -\cfrac{\sqrt{2}}{2} \right) \\ \quad \\ 3\theta= \begin{cases} \frac{3\pi }{4}\\ \frac{5\pi }{4} \end{cases}\qquad \theta= \begin{cases} \frac{\pi }{4}\\ \frac{5\pi }{12} \end{cases}\)
anonymous
  • anonymous
so..... I just have two solutions for #7, which are pi/4 and 5pi/12
jdoe0001
  • jdoe0001
yeap, due to the range being only \((0,2\pi)\)
anonymous
  • anonymous
ok then how do I do #8
jdoe0001
  • jdoe0001
cosine is negative only in the 3rd and 4th quadrants, thus
jdoe0001
  • jdoe0001
one sec on 8
anonymous
  • anonymous
wait one second, im going to show you how I solved it and you can tell me if its right
jdoe0001
  • jdoe0001
k
anonymous
  • anonymous
alright so for question #8, I only got 1 solution, 5pi/6
anonymous
  • anonymous
uhh.. Its hing to take me a long time to type an equation
anonymous
  • anonymous
let me just post a picture... give me one sec
jdoe0001
  • jdoe0001
ok
jdoe0001
  • jdoe0001
but \(\bf sin\left( \theta-\frac{\pi }{3} \right)+1=2\implies sin\left( \theta-\frac{\pi }{3} \right)=1 \\ \quad \\ \theta-\frac{\pi }{3} =sin^{-1}(1)\implies \theta-\cfrac{\pi }{3}=\cfrac{\pi }{2}\implies \theta=\cfrac{\pi }{2} +\cfrac{\pi }{3} \\ \quad \\ \theta=\cfrac{5\pi }{6}\)
jdoe0001
  • jdoe0001
so.. that one is ok how about 9)?
anonymous
  • anonymous
anonymous
  • anonymous
sorry I took so long, I was having trouble typing in the equation
anonymous
  • anonymous
im stuck on number 9 and 10
jdoe0001
  • jdoe0001
tis ok... hmm ok... hold the mayo on 9
anonymous
  • anonymous
what?
jdoe0001
  • jdoe0001
hmm hold the mayo, just tomatoes and mustard, one sec :)
anonymous
  • anonymous
ok...
jdoe0001
  • jdoe0001
do you recall what is \(\bf 1^2 = ?\)
anonymous
  • anonymous
yeah
jdoe0001
  • jdoe0001
k
anonymous
  • anonymous
1
jdoe0001
  • jdoe0001
\(\bf 3cos(\theta)+3=2sin^2(\theta) \\ \quad \\ 3[cos(\theta)+1]=2sin^2(\theta) \\ \quad \\ 3[cos(\theta)+1]=2[{\color{brown}{ 1-cos^2(\theta)}}] \\ \quad \\ 3[cos(\theta)+1]=2[{\color{brown}{ 1^2-cos^2(\theta)}}] \\ \quad \\ 3[cos(\theta)+1]=2[{\color{brown}{(1-cos(\theta))(1+cos(\theta))}}]\) folow it so far?
anonymous
  • anonymous
yeah
jdoe0001
  • jdoe0001
notice, \(1=1^2\), thus we use the difference of squares
anonymous
  • anonymous
alright makes sense
jdoe0001
  • jdoe0001
ok, one sec
jdoe0001
  • jdoe0001
\(\bf 3[cos(\theta)+1]=2[{(1-cos(\theta))(1+cos(\theta))}] \\ \quad \\ 3[cos(\theta)+1]=2[{(1-cos(\theta))(cos(\theta)+1)}] \\ \quad \\ 3[cos(\theta)+1]-2{(1-cos(\theta))(cos(\theta)+1)}=0 \\ \quad \\ {\color{brown}{ [cos(\theta)+1]}}[3-2(1-cos(\theta))]=0 \\ \quad \\ \begin{cases} cos(\theta)+1=0\implies cos(\theta)=-1\implies \theta=cos^{-1}(-1) \\ \quad \\ 3-2(1-cos(\theta))\implies 1-cos(\theta)=\cfrac{-3}{-2}\implies 1-\cfrac{3}{2}=cos(\theta) \end{cases}\) notice the red part, is the common factor
jdoe0001
  • jdoe0001
keeping in mind that \(cos(\theta)+1 \iff 1+cos(\theta)\)
jdoe0001
  • jdoe0001
thus \(\bf \begin{cases} \theta=cos^{-1}(-1)\\ 1-\cfrac{3}{2}=cos(\theta)\implies -\cfrac{1}{2}=cos(\theta)\implies cos^{-1}\left( -\cfrac{1}{2} \right)=\theta \end{cases}\)
anonymous
  • anonymous
so the solutions are just theta
jdoe0001
  • jdoe0001
cosine is -1 at \(\pi\) and -1/2 at the angles you found before, \(\cfrac{3\pi }{4},\cfrac{5\pi }{4}\)
anonymous
  • anonymous
woah woah woah, from where did we get 3pi/4 and 5pi/4
jdoe0001
  • jdoe0001
ehhe
jdoe0001
  • jdoe0001
from the 2nd case, the one where cosine is -1/2
anonymous
  • anonymous
so \[\cos^{-1} (\frac{ -1 }{ 2 })=\frac{ 3\pi }{ 4 },\frac{ 5\pi }{ 4 }\]
anonymous
  • anonymous
@jdoe0001 so those are the two solutions
jdoe0001
  • jdoe0001
yeap, plus the 1st case, where cosine was -1, which means \(\pi\)
anonymous
  • anonymous
so there are 3 solutions
anonymous
  • anonymous
pi,3pi/4,5pi/4
jdoe0001
  • jdoe0001
yeap
anonymous
  • anonymous
how about the last question
jdoe0001
  • jdoe0001
one sec
anonymous
  • anonymous
ok
jdoe0001
  • jdoe0001
\(\bf cos(2\theta)+3=5cos(\theta) \\ \quad \\ {\color{brown}{ 2cos^2(\theta)-1 }}+3=5cos(\theta) \\ \quad \\ 2cos^2(\theta)-1+3-5cos(\theta)=0 \\ \quad \\ 2cos^2(\theta)-5cos(\theta)+2=0\impliedby \textit{notice, is a quadratic}\)
jdoe0001
  • jdoe0001
was looking if it was factorable by integers, if not, then we'd do quadratic formula
jdoe0001
  • jdoe0001
yeap, is factorable
anonymous
  • anonymous
ok
jdoe0001
  • jdoe0001
\(\bf 2cos^2(\theta)-5cos(\theta)+2=0\implies [2cos(\theta)-1][cos(\theta)-2] \\ \quad \\ \begin{cases} 2cos(\theta)-1\implies cos(\theta)=\cfrac{1}{2}\implies \theta=cos^{-1}\left( \cfrac{1}{2} \right) \\ \quad \\ cos(\theta)-2\implies cos(\theta)=2\implies \theta=cos^{-1}(2) \end{cases}\)
jdoe0001
  • jdoe0001
and you can check the factoring with FOIL if you wish
anonymous
  • anonymous
yes it also works with FOIL
jdoe0001
  • jdoe0001
notice the 2nd case, it's angle whose cosine is 2 however cosine is always -1 or 1 or in between meanign the 2nd case can be tossed away, and only use the 1st one
jdoe0001
  • jdoe0001
and of course, cosine is 1/2 at \(\bf \cfrac{\pi }{3},\cfrac{5\pi }{3}\)
jdoe0001
  • jdoe0001
actually, had a missing 0 in the cases, but anyhow, they're = 0 =) \(\bf 2cos^2(\theta)-5cos(\theta)+2=0\implies [2cos(\theta)-1][cos(\theta)-2]=0 \\ \quad \\ \begin{cases} 2cos(\theta)-1=0\implies cos(\theta)=\cfrac{1}{2}\implies \theta=cos^{-1}\left( \cfrac{1}{2} \right) \\ \quad \\ cos(\theta)-2=0\implies cos(\theta)=2\implies \theta=cos^{-1}(2) \end{cases}\)
jdoe0001
  • jdoe0001
anyway, have to dash :)
anonymous
  • anonymous
WAIT SO WHAT ARE THE SOLUTIONS

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