# Rhts

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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

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