Question

Last question of my math homework! Finally it's been a long time! Help me and I'm done with math!

Answer 1

Solve please
\[\sin(2 \theta) - \cos \theta - 2 \sin \theta + 1 = 0\]

Answer 2

Steve: Didn't hear back from you regarding our discussion of your previously posted question. How did that turn out?

Answer 3

I figured it out and was right! I substituted the x back with 2 theta and solved.

Answer 4

Thanks for the help mathmale :)

Answer 5

All right. What are your thoughts regarding this new post? (You're welcome.)

Answer 6

This time, I think I can use the double angle formula since there is both sin and cos in the equation?

Answer 7

Chances are that this will work.
What have you done so far?

Answer 8

\[2\sin \theta \cos \theta - \cos \theta - 2\sin \theta + 1 = 0\]

Answer 9

I'm trying to factor this expression. Would you try that on your own, please? Is there a common factor in your equation?

Answer 10

I think you can distribute the 2sin theta, so now, I have\[2\sin \theta (\cos \theta - 1) - \cos \theta + 1 = 1\]

Answer 11

Oops, I typed that wrong, I meant = 0

Answer 12

What do you mean by "distribute the 2 sin theta"?

Answer 13

Sorry, factor out the 2 sin theta

Answer 14

Try it. I've been looking to factor out cos theta, by the way.

Answer 15

2sinθ(cosθ−1)−cosθ+1=0

Answer 16

Your format looks very similar to mine. Any idea of what to do next? Again I ask you whether your expression has a common factor.

Answer 17

Hmm, I realize I can cancel out the cosθ−1

Answer 18

with some manupulation you can make cos theta - 1 into 1 - cos theta.

Answer 19

Not "cancel out," Steve, but "factor out." If you drop the factor you've cancelled out, you 'll lose a root or two.

Answer 20

\[2\sin \theta (\cos \theta - 1) = \cos \theta - 1\]
\[2\sin \theta = \frac{ \cos \theta - 1 }{ \cos \theta - 1 }\]
\[2\sin \theta = 1\]
\[\sin \theta = \frac{ 1 }{ 2 }\]

Answer 21

I worked out the problem and that is what I got...

Answer 22

Not sure if it's right though

Answer 23

I've worked thru the same problem in a slightly different way and have come up with 3 solutions on the interval [0,pi].
How would you go about checking the validity of your one solution?

Answer 24

When solving an equation and coming up with a solution, never assume the solution is correct until after you've substituted it back into the original equation and find the equation to be true.

Answer 25

Oh okay.

Answer 26

So the two solutions I got are pi/6 and 5pi/6
How did you get 3 solutions?

Answer 27

2sinθ(cosθ−1)−cosθ+1=0
can be re-written as \[2\sin \theta (\cos \theta-1)-( ?? ) = 0\]

Answer 28

Steve, I asked whether you could factor the equation above. Have you been able to do that? If so, you'll end up with 2 factors, each of which you need to set = to zero separately and solve for theta. This is the procedure that got me 3 solutions on [0,pi].

Answer 29

ohhhh I see what you mean finally!

Answer 30

;)

Answer 31

Sorry to fail you, but I can't figure out the factors :(

Answer 32

We have \[2\sin \theta(\cos \theta - 1)-\cos \theta +1 = 0\]

Answer 33

and the 2nd half of this equation can be re-written to match that \[\cos \theta -1\]

Answer 34

factor on the left::

Answer 35

ohhhh, i am so stupid I'm sorry, I got this!

Answer 36

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

Answer 37

OK with this?

Answer 38

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

Answer 39

Can you now factor this latest equation?

Answer 40

You can try factoring your own equation also. That's the best way, probably, to determine whether your equation will "work" or not.

Mind factoring my last equation now?

Answer 41

My last equation definitely has a common factor. What is that factor?

Answer 42

I finally got it thanks so much for the 30 minutes of your life :)

Answer 43

It was well spent, Steve. Best of luck to you. Bye for now.