求函數f（x）在區間[-π/4，π/4]的最大值和最小值 f（x）=sin（2x+π/3）+sin（2x-π/3）+2cos²；x-1，x∈R

求函數f（x）在區間[-π/4，π/4]的最大值和最小值 f（x）=sin（2x+π/3）+sin（2x-π/3）+2cos²；x-1，x∈R

f（x）=sin2xcos（π/3）+cos2xsin（π/3）+sin2xcos（π/3）-cos2xsin（π/3）+cos2x
=2sin2xcos（π/3）+cos2x
=sin2x+cos2x
=√2sin（2x+π/4）
f（max）=√2
f（min）= -√2

Elden Ring Premiere

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