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Find any critical numbers of the function. (sin(x))^2 + cos(x) ) 0 < x < 2π
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Were you able to find the derivative function?
not sure if its correct : 2 (sin x) (cos x) - sin x ?
Yah looks good :)
ok then what? hehe
Critical points exist where lines tangent to our function are horizontal. Or in other words, when our derivative function is zero. \[\large 2\sin x \cos x - \sin x=0\] So we let our function equal zero like this, and solve for x.
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If we factor a sin x out of each term, it gives us,\[\large \sin x(2\cos x-1)=0\]right? :o
yupp i got it now i know what to do ! thanks !
ok cool :)
thanks!
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