1-4cos2x+2sinx=0, six<0, cosx<0

Question

1-4cos2x+2sinx=0, six<0, cosx<0</p>

Answer 1

1-4(1-2sin^2x)+2sinx=0

-3+8sin^2x+2sinx=0

sinx=t

8t^2+2t-3=0

t=1/8*(-1+-sqrt(1+24))=1/8*(-1+-5)

t1=1/2 t2=-3/4

 

sinx=-3/4

 x=П*(2n+1)+arcsin(-3/4)