PreCal(Please help)
posted by Abby .
Approximate the equation's soultions in the interval (o, 2pi). If possible find the exact solutions.
sin 2x sinx = cosx
I do not know where to start.

Expand sin 2x using the identity:
sin2x=2sin(x)cos(x)
We'll get:
2cos(x)sin²(x)=cos(x)
Transpose lefthandside to the right:
2cos(x)((1/2)sin²(x))=0
So
cos(x)=0, or
sin²(x)=1/2
Solve for x in (0,2π) as specified.
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