Ответы на вопрос:
Cosx=a 2a²+a-1=0 d=1+8=9 a1=(-1-3)/4=-1⇒cosa=-1⇒a=π+2πn,n∈z -7π/2≤π+2πn≤2π -7≤2+4n≤4 -9≤4n≤2 -9/4≤n≤1/2 n=-2⇒a=π-4π=-3π n=-1⇒a=π-2π=-π n=0⇒a=π a2=(-1+3)/4=1/2⇒cosx=1/2⇒x=π/3+2πk u x=-π/3+2πm -7π/2≤π/3+2πk≤2π -21≤2+12k≤12 -23≤12k≤10 -23/12≤k≤5/6 k=-1⇒a=π/3-2π=-5π/3 k=0⇒a=π/3 -7π/2≤-π/3+2πm≤2π -21≤-2+12m≤12 -19≤12m≤14 -19/12≤m≤7/6 m=-1⇒a=-π/3-2π=-7π/3 m=0⇒a=-π/3 m=1⇒a=-π/3+2π=5π/3
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