设向量a=(根号3sinx,cosx),向量b=(cosx,cosx),记f(x)=向量a×向量b(1)写出函数f(x)的最小正周期及递增区间;(2)若f(x0)=11/10,x0属于[π/4,π/2],求cos2x0的值
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![设向量a=(根号3sinx,cosx),向量b=(cosx,cosx),记f(x)=向量a×向量b(1)写出函数f(x)的最小正周期及递增区间;(2)若f(x0)=11/10,x0属于[π/4,π/2],求cos2x0的值](/uploads/image/z/4077471-39-1.jpg?t=%E8%AE%BE%E5%90%91%E9%87%8Fa%3D%28%E6%A0%B9%E5%8F%B73sinx%2Ccosx%29%2C%E5%90%91%E9%87%8Fb%3D%28cosx%2Ccosx%29%2C%E8%AE%B0f%28x%29%3D%E5%90%91%E9%87%8Fa%C3%97%E5%90%91%E9%87%8Fb%281%29%E5%86%99%E5%87%BA%E5%87%BD%E6%95%B0f%28x%29%E7%9A%84%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%E5%8F%8A%E9%80%92%E5%A2%9E%E5%8C%BA%E9%97%B4%EF%BC%9B%282%29%E8%8B%A5f%28x0%29%3D11%2F10%2Cx0%E5%B1%9E%E4%BA%8E%5B%CF%80%2F4%2C%CF%80%2F2%5D%2C%E6%B1%82cos2x0%E7%9A%84%E5%80%BC)
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