## onegirl 3 years ago 2 cos^2 x - 3 cos x + 1 = 0 for 0 is less than or equal to x is less than 2pi

1. dpaInc

the first term with the cosine... is it: $$\large 2cos^2x$$ or $$\large 2cos(2x)$$ i'm thinking it's the first one?

2. onegirl

yea the first one

3. dpaInc

think of it this way.. let y=cosx so your equation becomes: $$\large 2y^2-3y+1=0$$ can you solve this quadratic?

4. onegirl

yea i can

5. dpaInc

what is/are the solutions ? y = ???

6. z3529080

cos2x=2cos^2(x)-1

7. z3529080

2(2cos^2(x)-1)-3cosx+1=0 -> 4cos^2(x)-3cosx-1=0 -> (4cosx+1)(cosx-1)=0

8. onegirl

x1 = -0.5 x2 = -1

9. z3529080

so set 4cosx+1=0 or cos-1 =0

10. dpaInc

hmmm.. i got the same but POSITIVE values... i'm double checking...

11. dpaInc

yeah... they should be positive: $$\large 2y^2-3y+1=0$$ $$\large (2y-1)(y-1)=0$$ so 2y - 1= 0 gives y=1/2 y-1 = 0 gives y=1

12. dpaInc

ok so far?

13. onegirl

yea

14. z3529080

you forgot to x2

15. z3529080

it is 2cos2x

16. onegirl

it is 2 cos^2 x - 3

17. z3529080

oh i thought it is cos2x

18. dpaInc

since y=cosx, we have these two equations: $$\large cosx=\frac{1}{2}$$ and $$\large cosx=-1$$ can you solve these?

19. dpaInc

oops... that second one should be POSITIVE one...

20. dpaInc

do you use the unit circle?

21. onegirl

yes i do

22. joshi

this is probably a stupid question, but i'm a bit confused. you said let y = cosx so what happens to the first term 2cos2x? like what about the 2 infront of the x?

23. onegirl

@joshi its 2 cos^2 x -3 i made a mistake in typing it

24. dpaInc

ok... so look at what angle gives you a cosine of 1/2.... HINT: there are two of 'em from 0 to 2pi

25. joshi

ohhh ok that makes more sense haha

26. dpaInc

yeah... i got a clarification from the asker with my first question... :)

27. onegirl

60 degrees?

28. dpaInc

|dw:1350069794735:dw| yep... 60 degrees is one of 'em....

29. onegirl

or pi/3

30. dpaInc

you got the second angle?

31. onegirl

and 45 degrees?

32. dpaInc

nope... we're still solving cosx = 1/2 , right? x=60 degrees or pi/3 is one... the second solution is in the fourth quadrant...

33. dpaInc

|dw:1350070145499:dw|

34. dpaInc

or take a look at page 3: http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf

35. onegirl

240 degrees?

36. dpaInc

240 degrees is in the third quadrant... this is the fourth quadrant: |dw:1350070347242:dw|

37. onegirl

ohh ok 300 degrees..

38. dpaInc

yes... :) cos(60 degrees) = 1/2 and cos(300 degrees) = 1/2 now solve the other equation: cos x = 1

39. dpaInc

and remember, $$\large 0\color {red}{\le} x<2\pi$$

40. dpaInc

there should be only one solution for this one....

41. dpaInc

using the unit circle, what angle gives you an x-coordinate of 1?

42. onegirl

360 degrees?

43. dpaInc

yes but 2pi is not included in the interval you're looking at.. 360 degrees = 2pi so what other angle gives you an x-coordinate of 1?

44. dpaInc

HINT: $$\huge \le$$

45. onegirl

ohhh yea it will 180 degrees

46. dpaInc

no.. cos(180 degrees) = -1 try again....

47. dpaInc

cos x = 1 where $$\large 0\color {red}{\le} x<2\pi$$

48. onegirl

3pi/2?

49. dpaInc

what about the left side of $$\large 0\color {red}{\le} x<2\pi$$ ???? x = 0 ? cos 0 = ???

50. dpaInc

ok... let's put it this way... what's an angle that is coterminal with 360 degrees?

51. onegirl

positive 360 and negative -360

52. dpaInc

no.. the angle you want is 0 because cos(0 degrees) = 1 so you have 3 solutions: x=0, 60, 300 degrees but since the problem was stated in terms of radians, i'd convert those angles.

53. onegirl

so it will be x=1.823 radian