Question

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

Answer 1

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

Answer 2

remember sin2x=2sinxcosx
sin(x+x)=sinxcosx+sinxcosx
sin(2x)=2sinxcosx

Answer 3

thanks again, i have one more problem left :D