anonymous
  • anonymous
show that eq has atleast one root in the given interval cos2x+sinx=0 , [-pi/2,pi/2]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
1-2sin^2x+sinx=0 2sin^2x-sinx-1=0 2sinx(sinx-1) +1(sinx-1)=0 sinx=1 or sinx=-1/2 x=pi/2 or x=-pi/6
amistre64
  • amistre64
cos(2x) + sin(x) = 0 cos(2x) = sin(-x) cos^2 - sin^2 +sin(x) = 0 (1 - sin^2) +sin^2 + sin(x) = 0 1+sin(x) = 0 sin(x) = -1 x = -pi/2
amistre64
  • amistre64
ack!!!....

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amistre64
  • amistre64
really need to do these on paper ;)
anonymous
  • anonymous
\[cos(-\pi) = -1\] \[sin (-\pi/2) = -1\] \[cos(\pi) = -1\] \[sin(\pi/2) = 1\] So at least it will be 0 at \(\pi/2\). Another easy way would be if you found a positive and a negative then the intermediate value theorem would tell you you had a root somewhere in there.
anonymous
  • anonymous
It doesn't specifically ask you to find it

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