Question

1)sin2x=sinx
2)cosx+2cos^2x=0

Answer 1

1) sin2x=sinx
sin2x - sinx = 0
2sinx - sinx = 0
sinx (2cosx - 1) = 0
sinx = 0 => x = pik, kc Z
cosx = 1/2 => x = ± pi/3 + 2pik, kcZ

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2) cosx + 2cos^2x=0
cosx ( 1 + 2cosx) = 0
cosx = 0 => x = pi/2 + pik, kc Z
cosx = - 1/2 => x = ± 2pi/3 + 2pik, kcZ