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Whats your question

so you need to find the critical points of sin(x) you mean?

yes!

first plot y = sin(x)
do you know how to do that?

|dw:1352867080741:dw| somethin like that right?

yes good

:)

yes

orig equation.....find the absolute extrema of the fn on the closed interval
g(x)=2x+5cosx [0, 2pi]

I'm assuming you're in calculus right?

yes sir

oh that explains \(\sin(x)=\frac{2}{5}\)!!

yeah :}

You guys always write such through explanations :P . I wish I could get to your level one day : ) .

I hit
2nd ---> sin
then typed 2/5, closed the parenthesis and hit enter
This types in sin^-1(2/5)

0 to 2pi sorry.

it s ok :]

in this case, 0 < x < pi or 0 < x < 2pi gives the same number of solutions

Indeed. I just wanted to be Concise @jim_thompson5910 .

i gotcha

if you look at the unit circle
|dw:1352868694165:dw|

So that's why there are 2 solutions to sin(x) = 2/5 where 0 < x < 2pi

oooooooooooo okay i understand now ::}}}]]

ok great

okay gotchA!!!!!! thanks a million! :]