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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 :)
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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
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OpenStudy (anonymous):
lovely! thanks. have a wonderful night :)
myininaya (myininaya):
np
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