cos2x=cos²x-sin²x
Din formula fundamentala a trigonometriei (cos²x+sin²x=1) scoatem sin²x si il inlocuim formula lui cos2x.
=> sin²x=1-cos²x
=>cos2x=cos²x-1+cos²x=2cos²x-1
cos2x=2cos²x-1
Inlocuim
2cos²x-1+7cosx+4=0
2cos²x+7cosx+3=0
cosx=t, t∈[-1,1]
2t²+7t+3=0
Δ=49-4·2·3=49-24=25
t1=(-7+5)/4=-2/4=-1/2 ∈[-1, 1]
t2=(-7-5)/4=-12/4=-3 ∉ [-1, 1]
cosx=-1/2 => x ∈ {+/- arccos(-1/2) +2kπ l k ∈ Z}
x ∈ {+/- π/3 +2kπ | k ∈ Z}
