решите уравнение 2sin^2x+(2-корень2)соsx+корень2-2=0. укажите корни, принадлежащие отрезку [5п/2;7п/2]
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2-2сos²x+(2-√2)cosx+√2-2=0cosx=a
2a²-(2-√2)a-√2=0
D=4-4√2+2+8√2=4+4√2+2=(√2+2)²
a1=(2-√2-√2-2)/4=-√2/2⇒cosx=-√2/2⇒x=+-3π/4+2πn,n∈z
a2=(2-√2+√2+2)/4=1⇒cosx=1⇒x=2πk,k∈z