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anonymous
 5 years ago
show that eq has atleast one root in the given interval
cos2x+sinx=0 , [pi/2,pi/2]
anonymous
 5 years ago
show that eq has atleast one root in the given interval cos2x+sinx=0 , [pi/2,pi/2]

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.012sin^2x+sinx=0 2sin^2xsinx1=0 2sinx(sinx1) +1(sinx1)=0 sinx=1 or sinx=1/2 x=pi/2 or x=pi/6

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0cos(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
 5 years ago
Best ResponseYou've already chosen the best response.0really need to do these on paper ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[cos(\pi) = 1\] \[sin (\pi/2) = 1\] \[cos(\pi) = 1\] \[sin(\pi/2) = 1\] 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
 5 years ago
Best ResponseYou've already chosen the best response.0It doesn't specifically ask you to find it
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