Solve the given equation 2cos^2(theta)+sin(theta)=1

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Solve the given equation 2cos^2(theta)+sin(theta)=1

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

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

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