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need help of sin and cos

Mathematics
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Apr 8th, 2015

Transform the  equation: sin2 x = 1 + cos x;   1 – sin2 x + cos x = 0;

cos2 x + cos x = 0; cos x (cos x + 1) = 0

The resulting equation has two series of solutions:

if cos x = 0, then                         x = π/2 + 2π*n, where n is an integer (0, 1, –1, 2, –2, 3, –3, …)

if cos x = –1, then                       x = π + 2π*n, where n is an integer (0, 1, –1, 2, –2, 3, –3, …)


Apr 8th, 2015

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