Posted by Zinhle on Monday, March 18, 2013 at 3:24pm.
If 13sinx+5=0 and x[0;270],determine witout the use of a calculator,the value of sin2x

???  Writeacher, Monday, March 18, 2013 at 3:43pm
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Trig  Steve, Monday, March 18, 2013 at 3:53pm
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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