anonymous
  • anonymous
-2sin^2(x)-1cos(x)=-1 Looking for the smallest non-negative solution and the next smallest non-negative solution Smallest non-negative solution: 0, ? Next smallest non-negative solution: 360, ?
OCW Scholar - Single Variable Calculus
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Mani_Jha
  • Mani_Jha
\[2\sin ^{2}x+\cos x=1\] \[2-2\cos ^{2}x+cosx=1\] \[2\cos ^{2}x-cosx-1=0\] What do you get on solving this?
anonymous
  • anonymous
So far I have: -2sin^2(x)-cos(x)=-1 -2(1-cos^2(x))-cos(x)=-1 -2+2cos^2(x)-cos(x)=-1 2cos^2(x)-cos(x)-1=0 (cosx-1)(2cosx+1)=0 cosx=1 or cosx=-1/2 x=0,360,720,... or x=120, 240, ... therefore, Smallest non-negative solution=0 and the next smallest non-negative solution =120 But I've inputed: The smallest non-negative solution is: 0 (0.5/1) The next smallest non-negative solution is: 360 (0.5/1) The Number in the brackets are the points that I've received so it looks like I'm missing something numbers. I've tried putting 120 for the second number in the "next smallest non-negative solution" spot and it said that was wrong.
Mani_Jha
  • Mani_Jha
Maybe you should have excluded 0, because it can be both positive and negative. -0 is the same as 0. The smallest non-negative solution is: 120 The next smallest non-negative solution is: 360

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anonymous
  • anonymous
0 and all positives are non-negative.
anonymous
  • anonymous
0 2.094395.... 4.188790....
anonymous
  • anonymous
These are the first 3 non negative....
anonymous
  • anonymous
|dw:1332986306947:dw|
anonymous
  • anonymous
\[x=0\] \[x=\cos^{-1} \left( -1/2 \right)\]
anonymous
  • anonymous
first 2.
anonymous
  • anonymous
which gives the values mentioned earlier.
anonymous
  • anonymous
mani is right

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