Babynini
  • Babynini
Solve the given equation 2cos^2(theta)+sin(theta)=1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Babynini
  • Babynini
I've gotten up to 2sin^2(theta)+sin(theta)+1=0 but cann't figure out how to factor :/
Babynini
  • Babynini
(2sin(theta) ? ___)(sin(theta) ? ____)
freckles
  • 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: \[2 \sin^2(\theta)-\sin(\theta)-1=0\]

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Babynini
  • Babynini
ah yeah I didn't notice the 2sin^2theta was negative.
freckles
  • 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)
Babynini
  • Babynini
wait so now what do i do for factoring =.=
freckles
  • freckles
if you don't like factoring much you could use the quadratic formula
Babynini
  • Babynini
(2sin(theta) - 1)(sin(theta)+1) = 0 ?
Babynini
  • Babynini
I think factoring is usually nicer xD
Babynini
  • Babynini
dangit. that wouldn't give us the right answer.
freckles
  • 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)
freckles
  • freckles
so which where the -1 and 1 are
Babynini
  • Babynini
right.
freckles
  • freckles
switch *
Babynini
  • Babynini
huh? wouldn't that give the same answer?
freckles
  • freckles
\[2 \sin^2(\theta)-\sin(\theta)-1=0 \\ (2 \sin(\theta)+1)(\sin(\theta)-1)=0\]
Babynini
  • Babynini
which would be 2sin^2(theta)-2sin(theta)+sin(theta)-1=0 which is what we want yeah?
freckles
  • freckles
yep
Babynini
  • Babynini
I was confused about -2sin(theta) - sin(theta) = -sin(theta) but now I see hah
freckles
  • freckles
no no -2sin(theta)+sin(theta)=-sin(theta)
freckles
  • freckles
oh did you mean to write that
Babynini
  • Babynini
Shiz yeah, sorry.
Babynini
  • Babynini
Finals week man. Killing me.
Babynini
  • Babynini
K so now we have sin(theta)=1/2 and sin(theta)=1
freckles
  • freckles
almost
Babynini
  • Babynini
-1/2?
freckles
  • 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
Babynini
  • Babynini
K and sin = -1/2 at 5pi/6 and 7pi/6 ....right
freckles
  • 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
Babynini
  • Babynini
hm? it just says solve the given equation. No interval.
freckles
  • freckles
probably looking for all solutions then
Babynini
  • Babynini
so perhaps 5pi/6, 7pi/6, pi/2
freckles
  • freckles
ok but sin(5pi/6)=1/2 not -1/2
Babynini
  • Babynini
crap. 11pi/6
freckles
  • 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
freckles
  • freckles
and yes sin(pi/2)=1 so pi/2 is a solution to the equation sin(u)=1
freckles
  • freckles
but
freckles
  • 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
freckles
  • freckles
\[\theta=\frac{7\pi}{6}+2 \pi n \\ \theta=\frac{11\pi}{6}+2 \pi n \\ \theta=\frac{\pi}{2}+2 \pi n \\ \text{ where } n \text{ is integer }\]
Babynini
  • Babynini
Ah ok. I remember this :)

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