已知向量A=(1,cosx/2)向量B=(根3sinx/2+cosx/2,y)共线且有函数y=f(x)(1)若f(x)=1,x属于(0,2π)求x的值.(2)在△ABC中,角ABC的对边分别是abc且满足2acosC+c=2b,求函数f(B)的范围?
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![已知向量A=(1,cosx/2)向量B=(根3sinx/2+cosx/2,y)共线且有函数y=f(x)(1)若f(x)=1,x属于(0,2π)求x的值.(2)在△ABC中,角ABC的对边分别是abc且满足2acosC+c=2b,求函数f(B)的范围?](/uploads/image/z/5916716-44-6.jpg?t=%E5%B7%B2%E7%9F%A5%E5%90%91%E9%87%8FA%3D%281%2Ccosx%2F2%29%E5%90%91%E9%87%8FB%3D%28%E6%A0%B93sinx%2F2%2Bcosx%2F2%2Cy%29%E5%85%B1%E7%BA%BF%E4%B8%94%E6%9C%89%E5%87%BD%E6%95%B0y%3Df%28x%29%EF%BC%881%EF%BC%89%E8%8B%A5f%28x%29%3D1%2Cx%E5%B1%9E%E4%BA%8E%EF%BC%880%2C2%CF%80%EF%BC%89%E6%B1%82x%E7%9A%84%E5%80%BC.%EF%BC%882%EF%BC%89%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%2C%E8%A7%92ABC%E7%9A%84%E5%AF%B9%E8%BE%B9%E5%88%86%E5%88%AB%E6%98%AFabc%E4%B8%94%E6%BB%A1%E8%B6%B32acosC%2Bc%3D2b%2C%E6%B1%82%E5%87%BD%E6%95%B0f%28B%29%E7%9A%84%E8%8C%83%E5%9B%B4%3F)
已知向量A=(1,cosx/2)向量B=(根3sinx/2+cosx/2,y)共线且有函数y=f(x)(1)若f(x)=1,x属于(0,2π)求x的值.(2)在△ABC中,角ABC的对边分别是abc且满足2acosC+c=2b,求函数f(B)的范围?
已知向量A=(1,cosx/2)向量B=(根3sinx/2+cosx/2,y)共线且有函数y=f(x)
(1)若f(x)=1,x属于(0,2π)求x的值.
(2)在△ABC中,角ABC的对边分别是abc且满足2acosC+c=2b,求函数f(B)的范围?
已知向量A=(1,cosx/2)向量B=(根3sinx/2+cosx/2,y)共线且有函数y=f(x)(1)若f(x)=1,x属于(0,2π)求x的值.(2)在△ABC中,角ABC的对边分别是abc且满足2acosC+c=2b,求函数f(B)的范围?
则,有 A=kB (k≠0)
∴ (1,cos(x/2))=k(√3sin(x/2)+cos(x/2),y)
∴ 1/[√3sin(x/2)+cos(x/2)]=cos(x/2)/y
∴ y=√3sin(x/2)*cos(x/2)+cos²(x/2)
=(√3/2)*sinx+(1/2)*cosx+1/2
=sin(x+π/6)+1/2
所以,f(x)=sin(x+π/6)+1/2
(1) ∵ f(x)=1
∴ sin(x+π/6)+1/2=1
则 sin(x+π/6)=1/2
∵ x∈(0,2π)
∴ (x+π/6)∈(π/6,13π/6)
故 x+π/6=5π/6
∴ x=2π/3
(2) ∵ 2acosC+c=2b
∴ cosC=(2b-c)/2a
又 cosC=(b²+a²-c²)/2ab
∴ (2b-c)/2a=(b²+a²-c²)/2ab
则 2b²-bc=b²+a²-c²
∴ (b²+c²-a²)/2bc=1/2
即 cosA=1/2
∴ A=π/3
B+C=2π/3
∵B∈(0,2π/3)
∴π/6