anonymous
  • anonymous
Find all values of x in the interval [0,2pi] that satisfy the equation. 2sin^2x=1 and sin2x = cosx
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Psymon
  • Psymon
Well, for the top one you need to solve for sin(x) first, which includes getting rid of the 2nd power applied to it. Once we solve for sin(x) we can go from there.
anonymous
  • anonymous
Would you do that by \[\sqrt{\sin^2x}=\sqrt{1}\] square rooting both sides?
Psymon
  • Psymon
You forgor to divide by 2, that disappeared in your equation. But after you divide by 2 you would square root both sides, yes.

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anonymous
  • anonymous
so \[\sin x = \sqrt{1/2}\]
Psymon
  • Psymon
Dont forget plus/minus. We always have to have that if we take an even root of things.
anonymous
  • anonymous
Then after that what would be the next step? Or is that the final answer?
Psymon
  • Psymon
No, we actually need x. Solving for sinx doesnt cut it. So it looks best if we simplify what we have. Because once we simplify it, the value will match up with one of the very neat values that can be found on the unit circle: \[\pm \sqrt{\frac{ 1 }{ 2 }}=\pm \frac{ \sqrt{2} }{ 2 }\]Do you know your unit circle or have one handy to identify where sinx would equal this value for the positive and negative?
anonymous
  • anonymous
3pi /4
Psymon
  • Psymon
That would be one. There are 3 more actually. There are 2 angles for which sin = positive sqrt(2)/2 and 2 angles for which sin = negative sqrt(2)/2
anonymous
  • anonymous
Or would it be \[5\pi/4, \pi/4, and, 7\pi/4\]
Psymon
  • Psymon
Those would be the other 3. So yes, those are all 4 of your answers :3
anonymous
  • anonymous
Ok cool! And then for the other problem sin2x = cosx
Psymon
  • Psymon
Right. We need to use an identity. We cannot have our functions have different angles. One has a 2x and one has an x. We want that 2x to become an x by using a double-angle identity. Would you happen to know thedouble-angle formula for sin2x?
anonymous
  • anonymous
2 sinx cosx = cosx right?
Psymon
  • Psymon
Bingo. Now just divide both sides by cosx and then by 2.
anonymous
  • anonymous
Sinx = 1/2 ?
Psymon
  • Psymon
Yes. So there are two values in which sinx = 1/2 on your unit circle :3
anonymous
  • anonymous
\[\pi/3, and, 5\pi/3?\]
Psymon
  • Psymon
Nope. pi/3and 5pi/3 is where cos is 1/2
anonymous
  • anonymous
Oh cos is in the y right? \[5\pi/6\] and \[pi/6\]
Psymon
  • Psymon
sin is y, cos is x :P but yes, those are the two correct values :3
anonymous
  • anonymous
Would the steps be the same for inequalities? Like sin x \[\le 1/2\]
Psymon
  • Psymon
Hmm...well, we know that pi/6 and 5pi/6 = 1/2. So which interval keeps it le 1/2? Is it pi/6 to 5pi/6that is less than 1/2 or is it 5pi/6 to pi/6?
anonymous
  • anonymous
5pi/6 to pi/6

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