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anonymous
 one year ago
I have solved a trig problem and want to see if I got the correct answer.
anonymous
 one year ago
I have solved a trig problem and want to see if I got the correct answer.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so..what did you get for 7?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so each question has multiple solutions

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For question #7, the solutions I got were\[\pi/4,2\pi/3 \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and\[5\pi/12,2\pi/3n\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well. you may not need the 3n part, since you're only constrained to \((0,2\pi)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\bf 2cos(3\theta)+\sqrt{2}=0\implies 2cos(3\theta)=\sqrt{2} \\ \quad \\ cos(3\theta)=\cfrac{\sqrt{2}}{2}\implies 3\theta=cos^{1}\left( \cfrac{\sqrt{2}}{2} \right) \\ \quad \\ 3\theta= \begin{cases} \frac{3\pi }{4}\\ \frac{5\pi }{4} \end{cases}\qquad \theta= \begin{cases} \frac{\pi }{4}\\ \frac{5\pi }{12} \end{cases}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so..... I just have two solutions for #7, which are pi/4 and 5pi/12

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeap, due to the range being only \((0,2\pi)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok then how do I do #8

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0cosine is negative only in the 3rd and 4th quadrants, thus

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait one second, im going to show you how I solved it and you can tell me if its right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0alright so for question #8, I only got 1 solution, 5pi/6

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0uhh.. Its hing to take me a long time to type an equation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0let me just post a picture... give me one sec

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but \(\bf sin\left( \theta\frac{\pi }{3} \right)+1=2\implies sin\left( \theta\frac{\pi }{3} \right)=1 \\ \quad \\ \theta\frac{\pi }{3} =sin^{1}(1)\implies \theta\cfrac{\pi }{3}=\cfrac{\pi }{2}\implies \theta=\cfrac{\pi }{2} +\cfrac{\pi }{3} \\ \quad \\ \theta=\cfrac{5\pi }{6}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so.. that one is ok how about 9)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry I took so long, I was having trouble typing in the equation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0im stuck on number 9 and 10

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0tis ok... hmm ok... hold the mayo on 9

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmm hold the mayo, just tomatoes and mustard, one sec :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do you recall what is \(\bf 1^2 = ?\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\bf 3cos(\theta)+3=2sin^2(\theta) \\ \quad \\ 3[cos(\theta)+1]=2sin^2(\theta) \\ \quad \\ 3[cos(\theta)+1]=2[{\color{brown}{ 1cos^2(\theta)}}] \\ \quad \\ 3[cos(\theta)+1]=2[{\color{brown}{ 1^2cos^2(\theta)}}] \\ \quad \\ 3[cos(\theta)+1]=2[{\color{brown}{(1cos(\theta))(1+cos(\theta))}}]\) folow it so far?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0notice, \(1=1^2\), thus we use the difference of squares

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\bf 3[cos(\theta)+1]=2[{(1cos(\theta))(1+cos(\theta))}] \\ \quad \\ 3[cos(\theta)+1]=2[{(1cos(\theta))(cos(\theta)+1)}] \\ \quad \\ 3[cos(\theta)+1]2{(1cos(\theta))(cos(\theta)+1)}=0 \\ \quad \\ {\color{brown}{ [cos(\theta)+1]}}[32(1cos(\theta))]=0 \\ \quad \\ \begin{cases} cos(\theta)+1=0\implies cos(\theta)=1\implies \theta=cos^{1}(1) \\ \quad \\ 32(1cos(\theta))\implies 1cos(\theta)=\cfrac{3}{2}\implies 1\cfrac{3}{2}=cos(\theta) \end{cases}\) notice the red part, is the common factor

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0keeping in mind that \(cos(\theta)+1 \iff 1+cos(\theta)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thus \(\bf \begin{cases} \theta=cos^{1}(1)\\ 1\cfrac{3}{2}=cos(\theta)\implies \cfrac{1}{2}=cos(\theta)\implies cos^{1}\left( \cfrac{1}{2} \right)=\theta \end{cases}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so the solutions are just theta

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0cosine is 1 at \(\pi\) and 1/2 at the angles you found before, \(\cfrac{3\pi }{4},\cfrac{5\pi }{4}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0woah woah woah, from where did we get 3pi/4 and 5pi/4

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0from the 2nd case, the one where cosine is 1/2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so \[\cos^{1} (\frac{ 1 }{ 2 })=\frac{ 3\pi }{ 4 },\frac{ 5\pi }{ 4 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@jdoe0001 so those are the two solutions

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeap, plus the 1st case, where cosine was 1, which means \(\pi\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so there are 3 solutions

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how about the last question

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\bf cos(2\theta)+3=5cos(\theta) \\ \quad \\ {\color{brown}{ 2cos^2(\theta)1 }}+3=5cos(\theta) \\ \quad \\ 2cos^2(\theta)1+35cos(\theta)=0 \\ \quad \\ 2cos^2(\theta)5cos(\theta)+2=0\impliedby \textit{notice, is a quadratic}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0was looking if it was factorable by integers, if not, then we'd do quadratic formula

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\bf 2cos^2(\theta)5cos(\theta)+2=0\implies [2cos(\theta)1][cos(\theta)2] \\ \quad \\ \begin{cases} 2cos(\theta)1\implies cos(\theta)=\cfrac{1}{2}\implies \theta=cos^{1}\left( \cfrac{1}{2} \right) \\ \quad \\ cos(\theta)2\implies cos(\theta)=2\implies \theta=cos^{1}(2) \end{cases}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and you can check the factoring with FOIL if you wish

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes it also works with FOIL

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0notice 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and of course, cosine is 1/2 at \(\bf \cfrac{\pi }{3},\cfrac{5\pi }{3}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0actually, had a missing 0 in the cases, but anyhow, they're = 0 =) \(\bf 2cos^2(\theta)5cos(\theta)+2=0\implies [2cos(\theta)1][cos(\theta)2]=0 \\ \quad \\ \begin{cases} 2cos(\theta)1=0\implies cos(\theta)=\cfrac{1}{2}\implies \theta=cos^{1}\left( \cfrac{1}{2} \right) \\ \quad \\ cos(\theta)2=0\implies cos(\theta)=2\implies \theta=cos^{1}(2) \end{cases}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0anyway, have to dash :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0WAIT SO WHAT ARE THE SOLUTIONS
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