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Solve the equation on the interval [0, 2_).

cos x + 2 cos x sin x = 0

User Keyoke
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1 Answer

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cos x + 2 cos x sin x = 0, this is equivalent to cos x (1+ 2 sin x) = 0, so
we have cosx=0 or
1+ 2 sin x = 0,
cosx=0, x= arccos(o) = pi/2 + 2kpi
1+ 2 sin x = 0, sinx = - 1/2, implies x=arcsin(- 1/2) = -pi/6 +2kpi
the solution is x= pi/2 + 2kpi, or x =-pi/6 +2kpi, k is a relative integer
User Ojdo
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