已知函數(shù)f(x)=sin2x-2cosx+2cos2x.
(1)若f(x)=-1,求x的值;
(2)求f(x)的最大值和最小值.
【答案】
分析:(1)利用同角平方關(guān)系把函數(shù)化簡(jiǎn)為關(guān)于cosx的二次函數(shù),解出cosx的值,進(jìn)而解x
(2)對(duì)函數(shù)進(jìn)行配方可得,
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,結(jié)合二次函數(shù)及余弦函數(shù)的知識(shí)進(jìn)行求解
解答:解:(1)f(x)=1-cos
2x-2cosx+2(2cos
2x-1)=3cos
2x-2cosx-1=-1(1分)
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(2分)
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(2分)
(2)因?yàn)椋?img src="http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181948940904835/SYS201310241819489409048022_DA/3.png">,(4分)
所以,當(dāng)cosx=-1時(shí),f(x)
max=4 (2分)
當(dāng)
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時(shí),
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(2分)
點(diǎn)評(píng):本題主要考查了三角函數(shù)與二次函數(shù)的知識(shí)的綜合運(yùn)用,在求二次函數(shù)的最值時(shí),要注意-1≤cosx≤1的范圍與二次函數(shù)的對(duì)稱軸的關(guān)系