Question

(tgx-1)($sqrt{2}$sinx+1)=0 удовлетворяющие условию cosx<0

Answer 1

(tgx-1)(√2sinx+1)=0
cosx<0⇒x∈(π/2+2πn;3π/2+2πn)
tgx-1=0⇒tgx=1⇒x=π/4+πn +x∈(π/2+2πn;3π/2+2πn)⇒x=5π/4+2πn
√2sinx+1=0⇒sinx=-1/√2⇒x=(-1)^n*π/4+πn+x∈(π/2+2πn;3π/2+2πn)⇒x=5π/4+2πn U x=7π/4+2πn
Ответ x=5π/4+2πn U x=7π/4+2πn