Rhts
posted by Zinhle .
If 13sinx+5=0 and x[0;270],determine witout the use of a calculator,the value of sin2x

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13sinx+5=0
sinx = 5/13
sinx is negative in QIII,QIV, but we want the QIII value.
so, cosx = 12/13
sin2x = 2sinx cosx = 2(5/13)(12/13) = 120/169