Решить уравнение:

sin2x-cosx=0

2sinxcosx-cosx=0

cosx (2sinx-1) = 0

получим 2 случая

1) cosx=0

x=pi/2+pik, k∈Z

2) 2sinx-1=0

sinx=1/2

x=pi/6+2pik, k∈Z

x=5pi/6+2pik, k∈Z

sin (2x) - sin (x) = 0sin (2x) - cos (x) = 0

sin (2x) - sin (x) = - cos (x) = 0

sin (2x) + cos (x) = sin (x) / / cos (x) = 0

2sin (x) cos (x) - sin (x) = - cox (x) = 0

2sin (x/2) cos (x/2) (2cos (x) - 1) = - cos (x) = 0