F (x) = cosx + SiNx, it is proved that a belongs to (0,0.5 π) such that f (x + a) = f (x + 3a) is constant

F (x + a) = f (x + 3a) sin (x + a) + cos (x + a) = sin (x + 3a) + cos (x + 3a) sin (x + a) - sin (x + 3a) = cos (x + 3a) - cos (x + a) 2cos (x + 2a) sin (- a) = - 2Sin (x + 2a) sin (a) Sina [cos (x + 2a) - sin (x + 2a)] = 0sina = 0, because a belongs to (0, π / 2), then there is no

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