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

1. anonymous

2. anonymous

@jdoe0001

3. jdoe0001

so..what did you get for 7?

4. anonymous

so each question has multiple solutions

5. jdoe0001

right

6. anonymous

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

7. anonymous

and$5\pi/12,2\pi/3n$

8. jdoe0001

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

9. anonymous

ok

10. jdoe0001

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

11. anonymous

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

12. jdoe0001

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

13. anonymous

ok then how do I do #8

14. jdoe0001

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

15. jdoe0001

one sec on 8

16. anonymous

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

17. jdoe0001

k

18. anonymous

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

19. anonymous

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

20. anonymous

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

21. jdoe0001

ok

22. jdoe0001

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

23. jdoe0001

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

24. anonymous

25. anonymous

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

26. anonymous

im stuck on number 9 and 10

27. jdoe0001

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

28. anonymous

what?

29. jdoe0001

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

30. anonymous

ok...

31. jdoe0001

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

32. anonymous

yeah

33. jdoe0001

k

34. anonymous

1

35. 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?

36. anonymous

yeah

37. jdoe0001

notice, $$1=1^2$$, thus we use the difference of squares

38. anonymous

alright makes sense

39. jdoe0001

ok, one sec

40. 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

41. jdoe0001

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

42. jdoe0001

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

43. anonymous

so the solutions are just theta

44. jdoe0001

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

45. anonymous

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

46. jdoe0001

ehhe

47. jdoe0001

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

48. anonymous

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

49. anonymous

@jdoe0001 so those are the two solutions

50. jdoe0001

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

51. anonymous

so there are 3 solutions

52. anonymous

pi,3pi/4,5pi/4

53. jdoe0001

yeap

54. anonymous

55. jdoe0001

one sec

56. anonymous

ok

57. jdoe0001

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

58. jdoe0001

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

59. jdoe0001

yeap, is factorable

60. anonymous

ok

61. jdoe0001

$$\bf 2cos^2(\theta)-5cos(\theta)+2=0\implies [2cos(\theta)-1][cos(\theta)-2] \\ \quad \\ \begin{cases} 2cos(\theta)-1\implies cos(\theta)=\cfrac{1}{2}\implies \theta=cos^{-1}\left( \cfrac{1}{2} \right) \\ \quad \\ cos(\theta)-2\implies cos(\theta)=2\implies \theta=cos^{-1}(2) \end{cases}$$

62. jdoe0001

and you can check the factoring with FOIL if you wish

63. anonymous

yes it also works with FOIL

64. 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

65. jdoe0001

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

66. jdoe0001

actually, had a missing 0 in the cases, but anyhow, they're = 0 =) $$\bf 2cos^2(\theta)-5cos(\theta)+2=0\implies [2cos(\theta)-1][cos(\theta)-2]=0 \\ \quad \\ \begin{cases} 2cos(\theta)-1=0\implies cos(\theta)=\cfrac{1}{2}\implies \theta=cos^{-1}\left( \cfrac{1}{2} \right) \\ \quad \\ cos(\theta)-2=0\implies cos(\theta)=2\implies \theta=cos^{-1}(2) \end{cases}$$

67. jdoe0001

anyway, have to dash :)

68. anonymous

WAIT SO WHAT ARE THE SOLUTIONS