anonymous
  • anonymous
Solve for x: 2sin^2x=2 + cosx; -2π ≤ x ≤ 2π
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Psymon
  • Psymon
More identity tricks we have to use xD So we hve to chance that sin^2x because not only do we want the angle to be the same we, ifpossible, want to have the same trig function. So I want that sin^2x to become cosine something. The way this is done is using this identity: \[\sin ^{2}x+\cos ^{2}x=1\]So we want to solve for sin^2x in this identity.If I get sin^2x by itself, what do I have?
anonymous
  • anonymous
Could you possibly work it out by steps?
Psymon
  • Psymon
Well, all we want to do right now is use \[\sin ^{2}x+\cos ^{2}x= 1\]and our goal is to get sin^2x by itself. So all I meant to do was subtract both sides by cos^2x. That kinda make sense?

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anonymous
  • anonymous
Yes so it then becomes?
Psymon
  • Psymon
Well, I'm sure you could see that then I would have: \[\sin ^{2}x=1-\cos ^{2}x \]Right?
anonymous
  • anonymous
Yes but what happened to the 2 in front of the sin
Psymon
  • Psymon
I'm working on that. What I did above from theoriginal problem. This is something we have to do to solve it, though. All I needed to show you is that sin^2x is the exact same thing as (1-cos^2x). That was the whole point of this little detour. Now that I know this, I'm going to replace sin^2x in our problem. Meaning Ill have this: \[2(1-\cos ^{2}x)= 2 + cosx\] Kinda see what I did?
anonymous
  • anonymous
Yes I understand
Psymon
  • Psymon
Alright. Now what I just didnt isnt something crazy. Youll have to do this a lot. So I would definitely try to remember that you may often have to set up sin^2x + cos^2x = 1 and then solve for sin^2x or cos^2x. You use it a ton, so good thing to know. Okies, so now I simplify what we have. \[2-2\cos ^{2}x=2 + cosx\] Now notice we have a cos^2 a cos^1 anda number. This fits the format that we would have for a quadratic equation. Up til now, youre used to seeing something like: \[x ^{2}+2x + 1 = 0\]Well, this is the exact same thing, except we have cosines instead. So because of this, we have to follow the same rules. So the first thing we do is move evetrything to one side of the equal sign and haveit all set to 0. If I do this I get: \[2\cos ^{2}x+cosx = 0\] THat part make sense so far?
anonymous
  • anonymous
Yes
Psymon
  • Psymon
Okay, cool. So again, we want to treat this as if it were any other quadratic. So we solve it the same way we would solve \[2x ^{2}+x= 0\]So do you remember how you would go about factoring that and solving for x?
anonymous
  • anonymous
Umm yeah I think so
Psymon
  • Psymon
Alright, cool. So if you factor it correctly youll end up with 2 factors. Think you got them?
anonymous
  • anonymous
So it would be x(2x+1) then your factors would be 0 and -1/2
Psymon
  • Psymon
Correct. thats exactly what we do. Except this time we have cosx instead of x. SO what we actually hve is cosx = 0 cosx = -1/2 So we have to solve both of those now. There will be 4 answers, 2 for each factor.
anonymous
  • anonymous
How do we solve these?
Psymon
  • Psymon
We're looking at the unit circle again. THere are two angles where cosx = 0 and two angles where cosx = -1/2.
anonymous
  • anonymous
So 90 and 270 for the first and 120 and 240 for the second
Psymon
  • Psymon
Sounds good to me :3
anonymous
  • anonymous
Wow thanks your amazing
Psymon
  • Psymon
Glad ya got it :3

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