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

Question

anonymous · Answer

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 · Answer

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 · Answer

ack!!!....

amistre64 · Answer

really need to do these on paper ;)

anonymous · Answer

$$cos(-\pi) = -1$$
$$sin (-\pi/2) = -1$$
$$cos(\pi) = -1$$
$$sin(\pi/2) = 1$$

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 · Answer

It doesn't specifically ask you to find it