已知函数f(x)=sinx/2cosx/2+√3cos^2x/2 (1)求方程f(x)=0的解集;(2)如果△ABC的三边a,b,c满足b^2=ac,且边b所对的角为x,求角x的取值范围及此时函数f(x)的值域
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![已知函数f(x)=sinx/2cosx/2+√3cos^2x/2 (1)求方程f(x)=0的解集;(2)如果△ABC的三边a,b,c满足b^2=ac,且边b所对的角为x,求角x的取值范围及此时函数f(x)的值域](/uploads/image/z/8731051-43-1.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%3Dsinx%2F2cosx%2F2%EF%BC%8B%E2%88%9A3cos%EF%BC%BE2x%2F2+%281%29%E6%B1%82%E6%96%B9%E7%A8%8Bf%28x%29%3D0%E7%9A%84%E8%A7%A3%E9%9B%86%EF%BC%9B%EF%BC%882%EF%BC%89%E5%A6%82%E6%9E%9C%E2%96%B3ABC%E7%9A%84%E4%B8%89%E8%BE%B9a%2Cb%2Cc%E6%BB%A1%E8%B6%B3b%EF%BC%BE2%3Dac%2C%E4%B8%94%E8%BE%B9b%E6%89%80%E5%AF%B9%E7%9A%84%E8%A7%92%E4%B8%BAx%2C%E6%B1%82%E8%A7%92x%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4%E5%8F%8A%E6%AD%A4%E6%97%B6%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%E7%9A%84%E5%80%BC%E5%9F%9F)
已知函数f(x)=sinx/2cosx/2+√3cos^2x/2 (1)求方程f(x)=0的解集;(2)如果△ABC的三边a,b,c满足b^2=ac,且边b所对的角为x,求角x的取值范围及此时函数f(x)的值域
已知函数f(x)=sinx/2cosx/2+√3cos^2x/2 (1)求方程f(x)=0的解集;(2)如果△ABC的三边a,b,c
满足b^2=ac,且边b所对的角为x,求角x的取值范围及此时函数f(x)的值域
已知函数f(x)=sinx/2cosx/2+√3cos^2x/2 (1)求方程f(x)=0的解集;(2)如果△ABC的三边a,b,c满足b^2=ac,且边b所对的角为x,求角x的取值范围及此时函数f(x)的值域
1)
f(x)=sin(x/2)cos(x/2)+√3cos²(x/2)
=(sinx)/2+(√3cosx)/2-1/2
令cos(π/3)=1/2 sin(π/3)=√3/2
∴f(x)=sin(x+π/3)-1/2
f(x)=0
sin(x+π/3)=1/2
x+π/3=2kπ+π/6 →x=2kπ-π/6 ∪x+π/3=2kπ+2π/3→ x=2kπ+π/3
f(x)=0的解集{xix=2kπ-π/6 ∪ x=2kπ+π/3,k∈N+}
2)
由余弦定理得
cosx=(a²+c²-b²)/2ac=(a²+c²-ac)/2ac
由基本不等式得a²+c² ≥ 2ac
(a²+c²-ac)/2ac ≥ ac/2ac=1/2
所以cosx ≥ 1/2
∵x的取值范围是(0,π)
∴x的范围是(0,π/6)
当x的取值范围是(0,π/6)
f(x)=sin(x+π/3)-1/2
f(x)的值域为:(√3/2 - 1/2 ,1/2 )
嘿嘿,你题目么抄错把,那个sinx/2cosx/2???
1、f(x)=1/2sinx+√3/2cosx+√3/2 = sin(x+π/3)+√3/2 = 0
x = 2kπ -2π/3 (k∈Z)
2、cosx ≥ 1/2 ∴ 0 < x ≤ π/3
f(x)的值域为:[√3/2 - 1 ,√3/2 + 1]
f(x)=1\2sinx+√3\2cosx+√3\2
f(0)=1\2+√3\2