thanks

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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sin^2 = 1- cos ^2 ok? replace and let the terms all in left hand side , the right hand side =0. Show me that step, please
Do you know how to solve quadratic ?
no.....

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But we have no way to find out the solution without quadratic. How can I help??
a =pi
@cutegirl I shouldn't give you a free answer because it doesn't help you on long run.
got it
3cos a +3 = 1 - cos^2 a cos^2a +3cosa +2=0 (cos a+2) (cos a +1) =0 cos a +2 never =0 , because it is out of range hence , just cos a +1 =0 , then cos a =-1, hence a = pi + 2kpi.
yup. if limit is given, like from 0 to 2pi, then the answer is a =pi if you don't have any limit, then a = pi +2kpi
thanks so much !!!!
I don't know, you must look at your problem (the paper) , it is not up to me.
ow i c thanks

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