Question

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корень из (sinx*cosx)= - cosx

Answer 1

ОДЗ cosx≤0⇒x∈[π/2+2πn;3π/2+2πn]
sinxcosx=cos²x
cos²x-sinxcosx=0
cosx(cosx-sinx)=0
cosx=0⇒x=π/2+πn +ОДЗ⇒x=π/2 U x=3π/2
cosx-sinx=0/cosx≠0
1-tgx=0⇒tgx=1⇒x=π/4+πn +ОДЗ⇒х=5π/4