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Rezolvati ecuatia 2cosx+1=0, x∈[0,3π)

Răspuns :

Semaka2
2*cosx+1=0
cosx=-1/2
x=arccos(-1/2)
x=2π/3
solutia   generala
x=+/-2π/3+2kπ  
In  cazul  de  fata  
k=0  x=-2π/3<o  nu  se   accepta
k=0 x=+2π/3∈[0,3π)
k=1 x=-2π/3+2π=4π/3∈[0,3π)
k=1 x=2π/3+2π=8π/3∈[0,3π)
k=2 x=-2π/3+4π=10π/3>3π Ne   oprim  aici
x={2π/3;4π/3;8π/3}

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