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

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**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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