关于x的方程x²-xcosAcosB-cos(C/2)=0有一个根为1,则△ABC一定有()A.A=B B.B=C C.A=C D.A+B= π/2π打不出来可用a代替,详解,-)
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![关于x的方程x²-xcosAcosB-cos(C/2)=0有一个根为1,则△ABC一定有()A.A=B B.B=C C.A=C D.A+B= π/2π打不出来可用a代替,详解,-)](/uploads/image/z/122074-34-4.jpg?t=%E5%85%B3%E4%BA%8Ex%E7%9A%84%E6%96%B9%E7%A8%8Bx%26%23178%3B-xcosAcosB-cos%28C%2F2%29%3D0%E6%9C%89%E4%B8%80%E4%B8%AA%E6%A0%B9%E4%B8%BA1%2C%E5%88%99%E2%96%B3ABC%E4%B8%80%E5%AE%9A%E6%9C%89%EF%BC%88%EF%BC%89A.A%3DB+B.B%3DC+C.A%3DC+D.A%2BB%3D+%CF%80%2F2%CF%80%E6%89%93%E4%B8%8D%E5%87%BA%E6%9D%A5%E5%8F%AF%E7%94%A8a%E4%BB%A3%E6%9B%BF%2C%E8%AF%A6%E8%A7%A3%2C-%EF%BC%89)
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