cos2x - 5sinx - 3 = 0
[tex]cos 2x = 1-2sin^2x[/tex]
[tex]1-2sin^2x-5sinx-3=0[/tex]; Notam sin x cu a
[tex]2a^2 + 5a + 2 = 0 [/tex]
Δ = 9
[tex]a_1 = -2 [/tex]
sin x = -2 - Nu convine, sin fiind definit pe [-1;1]
[tex]a_2 = -\frac{1}{2} [/tex]
sin x = [tex]- \frac{1}{2} [/tex] => x = [tex](-1)^{k+1} * \frac{pi}{6} + kpi[/tex]
