Đáp án:
GTLNy=√3y=3 khi x=−π3+k2πx=−π3+k2π
GTNNy=−√3y=−3 khi x=2π3+k2πx=2π3+k2π (k∈Z)(k∈Z)
Giải thích các bước giải:
y=sin(x−2π3)−sinxy=sin(x−2π3)−sinx
=−12sinx−√32cosx−sinx=−12sinx−32cosx−sinx
=−32sinx−√32cosx=−32sinx−32cosx
=−√3(√32sinx−12cosx)=−3(32sinx−12cosx)
=−√3sin(x−π6)=−3sin(x−π6)
Do −1≤sin(x−π6)≤1−1≤sin(x−π6)≤1 ∀x∀x
⇒√3≥−√3sin(x−π6)≥−√3⇒3≥−3sin(x−π6)≥−3
⇒√3≥y≥−√3⇒3≥y≥−3
Vậy GTLNy=√3y=3 khi sin(x−π6)=−1⇔x−π6=−π2+k2πsin(x−π6)=−1⇔x−π6=−π2+k2π (k∈Z)(k∈Z)
⇔x=−π3+k2π⇔x=−π3+k2π
GTNNy=−√3y=−3 khi sin(x−π6)=1⇔x−π6=π2+k2πsin(x−π6)=1⇔x−π6=π2+k2π (k∈Z)(k∈Z)
⇔x=2π3+k2π⇔x=2π3+k2π (k∈Z)(k∈Z).
