已知函数 f(x)=1- 3 sin2x+2co s 2 x ．（1）求f（x）的最大值及取得最大值时的x集合；（ 设函数 f(x)=2co s 2 x+ ...
\u5df2\u77e5\u51fd\u6570 f(x)= 3 sin2x+2co s 2 x-m \uff0e\uff081\uff09\u82e5\u65b9\u7a0bf\uff08x\uff09=0\u5728 x\u2208[0\uff0c \u03c0 2 ]\uff081\uff09 f(x)=2sin(2x+ \u03c0 6 )+1-m \uff0c\u2234 m=2sin(2x+ \u03c0 6 )+1 \u5728 [0\uff0c \u03c0 2 ] \u5185\u6709\u89e3\u2235 0\u2264x\u2264 \u03c0 2 \uff0c\u2234 \u03c0 6 \u22642x+ \u03c0 6 \u2264 7\u03c0 6 \u2234 0\u22642sin(2x+ \u03c0 6 )+1\u22643 \uff0c\u22340\u2264m\u22643\uff082\uff09\u2235m=3\uff0c\u2234 f(A)=2sin(2A+ \u03c0 6 )-2=-1 \uff0c\u2234 sin(2A+ \u03c0 6 )= 1 2 \uff0c\u2234 2A+ \u03c0 6 = \u03c0 6 +2k\u03c0 \u6216 2A+ \u03c0 6 = 5\u03c0 6 +2k\u03c0\uff0c(k\u2208Z) \u2235A\u2208\uff080\uff0c\u03c0\uff09\u2234 A= \u03c0 3 \u2235 b+c=2\u22652 bc \u5f53\u4e14\u4ec5\u5f53b=c\u65f6bc\u6709\u6700\u5927\u503c1\uff0e\u2235a 2 =b 2 +c 2 -2bccosA=\uff08b+c\uff09 2 -3bc=4-3bc\uff0c\u2234a\u6709\u6700\u5c0f\u503c1\uff0c\u6b64\u65f6b=c=1\uff0e
\u6c42\u5bfc\u5f97\uff1af\u2032\uff08x\uff09=-4sinxcosx+2 3 cos2x=-2sin2x+2 3 cos2x=4sin\uff08 \u03c0 3 -2x\uff09\uff0c\u4ee4f\u2032\uff08x\uff09=0\uff0c\u5f97\u5230x= \u03c0 6 \uff0c\u2235f\uff080\uff09=2+a\uff0cf\uff08 \u03c0 2 \uff09=a\uff0cf\uff08 \u03c0 6 \uff09=3+a\uff0c\u2234\u51fd\u6570\u7684\u6700\u5c0f\u503c\u4e3aa\uff0c\u53c8\u51fd\u6570\u533a\u95f4 [0\uff0c \u03c0 2 ] \u4e0a\u7684\u6700\u5c0f\u503c\u4e3a-4\uff0c\u5219a=-4\uff0e\u6545\u7b54\u6848\u4e3a\uff1a-4
| （本小题满分14分） （1）f（x）=1-
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