已知函数 f(x)= 3 sin2x+2co s 2 x-m .(1)若方程f(x)=0在 x∈[0, π 2 ]
已知函数f(x)=2cos2x+3sin2x.(1)求f(x...
\u5df2\u77e5\u51fd\u6570f(x)=3sin(2x+\u03c0/6\uff09\u82e5x0\u2208[0,2\u03c0),\u4e14f(x0)=3/2,\u6c42x0\u7684
\uff081\uff09f(x0)=3/2\uff0c\u6240\u4ee5sin(2x+\u03c0/6)=1/2.
2x0+\u03c0/6=2k\u03c0+\u03c0/6\u62162k\u03c0+5\u03c0/6\uff0ck\u2208Z.
\u56e0\u4e3ax0\u2208[0,2\u03c0)\uff0c\u6240\u4ee5x0=0\u6216\u03c0\u6216\u03c0/3\u62164\u03c0/3.
\uff082\uff09f(x)=2sin2(x+\u03c0/12).
\u03c0/12-m=k\u03c0+\u03c0/2\uff0ck\u2208Z
\u6700\u5c11\u5411\u53f3\u5e73\u79fb7\u03c0/12\u4e2a\u5355\u4f4d\u3002
\uff083\uff09f(x)=3sin(2x+\u03c0/6)=a
\u56e0\u4e3ax\u2208[0,\u03c0/2)\uff0c\u6240\u4ee52x+\u03c0/6\u2208[\u03c0/6,7\u03c0/6)
\u4f7ff(x)=a\u4ec5\u6709\u4e00\u4e2a\u5b9e\u6570\u89e3\uff0c\u52193sin(7\u03c0/6)<a<3sin(5\u03c0/6)
\u89e3\u5f97-3/2<a<3/2.
\uff081\uff09\u5316\u7b80\u53ef\u5f97f\uff08x\uff09=2cos2x+3sin 2x=1+cos2 x+3sin 2x=1+2\uff0812cos 2x+3sin 2x\uff09=2sin\uff082x+\u03c06\uff09+1\uff0c\u2234f\uff08x\uff09\u7684\u6700\u5c0f\u6b63\u5468\u671fT=\u03c0\uff0c\u4ee42k\u03c0-\u03c02\u22642x+\u03c06\u22642k\u03c0+\u03c02\uff0ck\u2208Z\uff0c\u89e3\u5f97k\u03c0-\u03c03\u2264x\u2264k\u03c0+\u03c06\uff0cf\uff08x\uff09\u7684\u5355\u8c03\u9012\u589e\u533a\u95f4\u4e3a[k\u03c0-\u03c03\uff0ck\u03c0+\u03c06\uff0c\uff08k\u2208Z\uff09\uff0e\uff082\uff09\u2235x\u2208[-\u03c04\uff0c\u03c04]\uff0c\u22342x+\u03c06\u2208[?\u03c03\uff0c2\u03c03]\uff0c\u2234sin\uff082x+\u03c06\uff09\u2208[?3<tr
(1) f(x)=2sin(2x+ )+1-m ,∴ m=2sin(2x+ )+1 在 [0, ] 内有解 ∵ 0≤x≤ ,∴ ≤2x+ ≤ ∴ 0≤2sin(2x+ )+1≤3 ,∴0≤m≤3
(2)∵m=3,∴ f(A)=2sin(2A+ )-2=-1 , ∴ sin(2A+ )= ,∴ 2A+ = +2kπ 或 2A+ = +2kπ,(k∈Z) ∵A∈(0,π)∴ A= ∵ b+c=2≥2 |