## Babynini one year ago Solve the given equation 2cos^2(theta)+sin(theta)=1

1. Babynini

I've gotten up to 2sin^2(theta)+sin(theta)+1=0 but cann't figure out how to factor :/

2. Babynini

(2sin(theta) ? ___)(sin(theta) ? ____)

3. freckles

testing what you said for my ownself: $2(1-\sin^2(\theta))+\sin(\theta)-1=0 \\ \text{ by pythagorean identitiy; also subtract 1 on both sides } \\ 2-2\sin^2(\theta)+\sin(\theta)-1=0 \\ \text{ by distribute property } \\ -2\sin^2(\theta)+\sin(\theta)+1=0 \text{ combining like terms }$ Think you left a sign off on the first term here. I'm also going to multiply -1 on both sides: $2 \sin^2(\theta)-\sin(\theta)-1=0$

4. Babynini

ah yeah I didn't notice the 2sin^2theta was negative.

5. freckles

anyways if you are doing trial factors you don't have a lot of trials here since -1 is 1(-1) or -1(1)

6. Babynini

wait so now what do i do for factoring =.=

7. freckles

if you don't like factoring much you could use the quadratic formula

8. Babynini

(2sin(theta) - 1)(sin(theta)+1) = 0 ?

9. Babynini

I think factoring is usually nicer xD

10. Babynini

dangit. that wouldn't give us the right answer.

11. freckles

well if you want to do factoring if you multiply that out the middle term will be 2sin(theta)-sin(theta which will not be -sin(theta)

12. freckles

so which where the -1 and 1 are

13. Babynini

right.

14. freckles

switch *

15. Babynini

huh? wouldn't that give the same answer?

16. freckles

$2 \sin^2(\theta)-\sin(\theta)-1=0 \\ (2 \sin(\theta)+1)(\sin(\theta)-1)=0$

17. Babynini

which would be 2sin^2(theta)-2sin(theta)+sin(theta)-1=0 which is what we want yeah?

18. freckles

yep

19. Babynini

I was confused about -2sin(theta) - sin(theta) = -sin(theta) but now I see hah

20. freckles

no no -2sin(theta)+sin(theta)=-sin(theta)

21. freckles

oh did you mean to write that

22. Babynini

Shiz yeah, sorry.

23. Babynini

Finals week man. Killing me.

24. Babynini

K so now we have sin(theta)=1/2 and sin(theta)=1

25. freckles

almost

26. Babynini

-1/2?

27. freckles

$2 \sin^2(\theta)-\sin(\theta)-1=0 \\ (2 \sin(\theta)+1)(\sin(\theta)-1)=0 \\ \sin(\theta)=\frac{-1}{2} \text{ or } \sin(\theta)=1$ last equation right just a sign off on the first

28. Babynini

K and sin = -1/2 at 5pi/6 and 7pi/6 ....right

29. freckles

can I ask if you want me to check your answers can you give me the intervals in which you want to solve them

30. Babynini

hm? it just says solve the given equation. No interval.

31. freckles

probably looking for all solutions then

32. Babynini

so perhaps 5pi/6, 7pi/6, pi/2

33. freckles

ok but sin(5pi/6)=1/2 not -1/2

34. Babynini

crap. 11pi/6

35. freckles

sin(7pi/6)=-1/2 so 7pi/6 is a solution sin(11pi/6)=-1/2 so 11pi/6 is a solution yep yep

36. freckles

and yes sin(pi/2)=1 so pi/2 is a solution to the equation sin(u)=1

37. freckles

but

38. freckles

if they want all the solutions and since we are working with sin and cos just +2pi*n and say where n is an integer

39. freckles

$\theta=\frac{7\pi}{6}+2 \pi n \\ \theta=\frac{11\pi}{6}+2 \pi n \\ \theta=\frac{\pi}{2}+2 \pi n \\ \text{ where } n \text{ is integer }$

40. Babynini

Ah ok. I remember this :)