大致方法如下,但是可能會(huì)有計(jì)算錯(cuò)誤,僅供參考
1,∵√sin2x ,∴sin2x≥0 ,∴得kπ≤x≤2kπ+π/2.（k∈Z） 又∵4-x²≥0,∴-2≤x≤2.∴綜上,｛x|-2≤x≤-π/2,或0≤x≤π/2}
2.f(x)=sin(x-2Φ),令g(t)=sint,則g(t)的對(duì)稱軸為t=kπ+π/2,∴當(dāng)x-2Φ=kπ+π/2時(shí),即x=kπ+π/2+2Φ為ƒ(x)的對(duì)稱軸,∴當(dāng)x=0時(shí),解得φ=π/4-kπ/2.
3.由題可知,T=(7π/12-π/12)Χ2=π,∴ω=2π/T=2,且當(dāng)x=π/12時(shí),函數(shù)的最大值為3,當(dāng)x=7π/12時(shí)函數(shù)最小值為-3,代入得：A=3,φ=π/3.所以,解析式為：y=3sin(2x+π/3) 經(jīng)驗(yàn)證,符合題意