Solve in the interval (0, 2pi)?

sin2x=2sinx*cosx

2sinx*cosx=√2cosx

cosx(2sinx-√2)=0

1st case cosx=0

=> x=(2k+1)pi/2, k integer., from (0, 2pi):

pi/2, 3pi/2.

2nd case 2sinx-√2=0

=> sinx=√2/2

x=(-1)^k*pi/4+kpi, from (0, 2pi)::

pi/4, 3pi/4

sin2x= (√2)cosx

sin(2x) = 2sin(x)cos(x) = (√2)cos(x)

2sin(x) = (√2)

sin(x) = (√2)/2 .... x = π/4 = 45°

sin(x) is positive in the 1st and 4th quadrant

therefore x=(π/4) and x=2π-(π/4)

Hint:

write sin(2x) as 2sin(x)cos(x)

and don't forget to find both solutions