Question

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

Answer 1

@jdoe0001

Answer 2

so..what did you get for 7?

Answer 3

so each question has multiple solutions

Answer 4

right

Answer 5

For question #7, the solutions I got were\[\pi/4,2\pi/3 \]

Answer 6

and\[5\pi/12,2\pi/3n\]

Answer 7

well. you may not need the 3n part, since you're only constrained to \((0,2\pi)\)

Answer 8

ok

Answer 9

\(\bf 2cos(3\theta)+\sqrt{2}=0\implies 2cos(3\theta)=-\sqrt{2}
\\ \quad \\
cos(3\theta)=-\cfrac{\sqrt{2}}{2}\implies 3\theta=cos^{-1}\left( -\cfrac{\sqrt{2}}{2} \right)
\\ \quad \\
3\theta=
\begin{cases}
\frac{3\pi }{4}\\
\frac{5\pi }{4}
\end{cases}\qquad \theta=
\begin{cases}
\frac{\pi }{4}\\
\frac{5\pi }{12}
\end{cases}\)

Answer 10

so..... I just have two solutions for #7, which are pi/4 and 5pi/12

Answer 11

yeap, due to the range being only \((0,2\pi)\)

Answer 12

ok then how do I do #8

Answer 13

cosine is negative only in the 3rd and 4th quadrants, thus

Answer 14

one sec on 8

Answer 15

wait one second, im going to show you how I solved it and you can tell me if its right

Answer 16

k

Answer 17

alright so for question #8, I only got 1 solution, 5pi/6

Answer 18

uhh.. Its hing to take me a long time to type an equation

Answer 19

let me just post a picture... give me one sec

Answer 20

ok

Answer 21

but \(\bf sin\left( \theta-\frac{\pi }{3} \right)+1=2\implies sin\left( \theta-\frac{\pi }{3} \right)=1
\\ \quad \\
\theta-\frac{\pi }{3} =sin^{-1}(1)\implies \theta-\cfrac{\pi }{3}=\cfrac{\pi }{2}\implies \theta=\cfrac{\pi }{2}
+\cfrac{\pi }{3}
\\ \quad \\
\theta=\cfrac{5\pi }{6}\)

Answer 22

so.. that one is ok
how about 9)?

Answer 23

sorry I took so long, I was having trouble typing in the equation

Answer 24

im stuck on number 9 and 10

Answer 25

tis ok... hmm ok... hold the mayo on 9

Answer 26

what?

Answer 27

hmm hold the mayo, just tomatoes and mustard, one sec :)

Answer 28

ok...

Answer 29

do you recall what is \(\bf 1^2 = ?\)

Answer 30

yeah

Answer 31

k

Answer 32

1

Answer 33

\(\bf 3cos(\theta)+3=2sin^2(\theta)
\\ \quad \\
3[cos(\theta)+1]=2sin^2(\theta)
\\ \quad \\
3[cos(\theta)+1]=2[{\color{brown}{ 1-cos^2(\theta)}}]
\\ \quad \\
3[cos(\theta)+1]=2[{\color{brown}{ 1^2-cos^2(\theta)}}]
\\ \quad \\
3[cos(\theta)+1]=2[{\color{brown}{(1-cos(\theta))(1+cos(\theta))}}]\)

folow it so far?

Answer 34

yeah

Answer 35

notice, \(1=1^2\), thus we use the difference of squares

Answer 36

alright makes sense

Answer 37

ok, one sec

Answer 38

\(\bf 3[cos(\theta)+1]=2[{(1-cos(\theta))(1+cos(\theta))}]
\\ \quad \\
3[cos(\theta)+1]=2[{(1-cos(\theta))(cos(\theta)+1)}]
\\ \quad \\
3[cos(\theta)+1]-2{(1-cos(\theta))(cos(\theta)+1)}=0
\\ \quad \\
{\color{brown}{ [cos(\theta)+1]}}[3-2(1-cos(\theta))]=0
\\ \quad \\
\begin{cases}
cos(\theta)+1=0\implies cos(\theta)=-1\implies \theta=cos^{-1}(-1)
\\ \quad \\
3-2(1-cos(\theta))\implies 1-cos(\theta)=\cfrac{-3}{-2}\implies 1-\cfrac{3}{2}=cos(\theta)
\end{cases}\)

notice the red part, is the common factor

Answer 39

keeping in mind that \(cos(\theta)+1 \iff 1+cos(\theta)\)

Answer 40

thus \(\bf \begin{cases}
\theta=cos^{-1}(-1)\\
1-\cfrac{3}{2}=cos(\theta)\implies -\cfrac{1}{2}=cos(\theta)\implies cos^{-1}\left( -\cfrac{1}{2} \right)=\theta
\end{cases}\)

Answer 41

so the solutions are just theta

Answer 42

cosine is -1 at \(\pi\)
and -1/2 at the angles you found before, \(\cfrac{3\pi }{4},\cfrac{5\pi }{4}\)

Answer 43

woah woah woah, from where did we get 3pi/4 and 5pi/4

Answer 44

ehhe

Answer 45

from the 2nd case, the one where cosine is -1/2

Answer 46

so \[\cos^{-1} (\frac{ -1 }{ 2 })=\frac{ 3\pi }{ 4 },\frac{ 5\pi }{ 4 }\]

Answer 47

@jdoe0001 so those are the two solutions

Answer 48

yeap, plus the 1st case, where cosine was -1, which means \(\pi\)

Answer 49

so there are 3 solutions

Answer 50

pi,3pi/4,5pi/4

Answer 51

yeap

Answer 52

how about the last question

Answer 53

one sec

Answer 54

ok

Answer 55

\(\bf cos(2\theta)+3=5cos(\theta)
\\ \quad \\
{\color{brown}{ 2cos^2(\theta)-1 }}+3=5cos(\theta)
\\ \quad \\
2cos^2(\theta)-1+3-5cos(\theta)=0
\\ \quad \\
2cos^2(\theta)-5cos(\theta)+2=0\impliedby \textit{notice, is a quadratic}\)

Answer 56

was looking if it was factorable by integers, if not, then we'd do quadratic formula

Answer 57

yeap, is factorable

Answer 58

ok

Answer 59

\(\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}\)

Answer 60

and you can check the factoring with FOIL if you wish

Answer 61

yes it also works with FOIL

Answer 62

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

Answer 63

and of course, cosine is 1/2 at \(\bf \cfrac{\pi }{3},\cfrac{5\pi }{3}\)

Answer 64

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}\)

Answer 65

anyway, have to dash :)

Answer 66

WAIT SO WHAT ARE THE SOLUTIONS