Mathematics OpenStudy (anonymous):

Find the exact solutions to each equation for the interval 0<=x<2pi. sin 2x=-sqrt2 cos x myininaya (myininaya):

sin2x=2sinxcosx 2sinxcosx=-sqrt{2}cosx 2sinxcosx+sqrt{2}cosx=0 cosx(2sinx+sqrt{2}=0 cosx=0 => x=pi/2, 3pi/2 2sinx+sqrt{2}=0 2sinx=-sqrt{2} sinx=-sqrt{2}/2 =>x=5pi/4, 7pi/4 myininaya (myininaya):

remember sin2x=2sinxcosx sin(x+x)=sinxcosx+sinxcosx sin(2x)=2sinxcosx OpenStudy (anonymous):

thanks again, i have one more problem left :D

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