已知函数f(x)=sin(2x+π/6)+1/2在△ABC中,角A,B,C所对的边分别为a,b,c.已知1/(2sinA-sinC)=cosB/(sinBcosC),且f(A)=(1+√3)/2,求∠C的大小
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![已知函数f(x)=sin(2x+π/6)+1/2在△ABC中,角A,B,C所对的边分别为a,b,c.已知1/(2sinA-sinC)=cosB/(sinBcosC),且f(A)=(1+√3)/2,求∠C的大小](/uploads/image/z/7172037-45-7.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dsin%282x%2B%CF%80%2F6%29%2B1%2F2%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%2C%E8%A7%92A%2CB%2CC%E6%89%80%E5%AF%B9%E7%9A%84%E8%BE%B9%E5%88%86%E5%88%AB%E4%B8%BAa%2Cb%2Cc.%E5%B7%B2%E7%9F%A51%2F%282sinA-sinC%29%3DcosB%2F%EF%BC%88sinBcosC%EF%BC%89%2C%E4%B8%94f%28A%29%3D%281%2B%E2%88%9A3%29%2F2%2C%E6%B1%82%E2%88%A0C%E7%9A%84%E5%A4%A7%E5%B0%8F)
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