Hannah_Ahn
Solve algebraically. Express the roots in exact form. 2sinxcosx1 = 0
the answer is pi over 4 + n pi
and I don't know the steps
please help,
thanks :)



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

myininaya
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when is sin(u), 1?

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

Hannah_Ahn
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pi over 2

myininaya
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to solve for x just divide both sides by 2

Hannah_Ahn
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lovely!
thanks.
have a wonderful night :)

myininaya
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np