Question

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

Answer 1

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

Answer 2

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

Answer 3

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:
\[2 \sin^2(\theta)-\sin(\theta)-1=0\]

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

I think factoring is usually nicer xD

Answer 10

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

Answer 11

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)

Answer 12

so which where the -1 and 1 are

Answer 13

right.

Answer 14

switch *

Answer 15

huh? wouldn't that give the same answer?

Answer 16

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

Answer 17

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

Answer 18

yep

Answer 19

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

Answer 20

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

Answer 21

oh did you mean to write that

Answer 22

Shiz yeah, sorry.

Answer 23

Finals week man. Killing me.

Answer 24

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

Answer 25

almost

Answer 26

-1/2?

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

probably looking for all solutions then

Answer 32

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

Answer 33

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

Answer 34

crap. 11pi/6

Answer 35

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

Answer 36

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

Answer 37

but

Answer 38

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

Answer 39

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

Answer 40

Ah ok. I remember this :)