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Mathematics 22 Online
OpenStudy (anonymous):

Solve algebraically. Express the roots in exact form. 2sinxcosx-1 = 0 the answer is pi over 4 + n pi and I don't know the steps please help, thanks :)

myininaya (myininaya):

sin(2x)-1=0 sin(2x)=1

myininaya (myininaya):

when is sin(u), 1?

myininaya (myininaya):

when u=pi/2+2npi u=3pi/2+2npi so that means 2x=pi/2+2npi 2x=3pi/2+2npi

OpenStudy (anonymous):

pi over 2

myininaya (myininaya):

to solve for x just divide both sides by 2

OpenStudy (anonymous):

lovely! thanks. have a wonderful night :)

myininaya (myininaya):

np

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