How would you solve this???

$\displaystyle

3sin(6x)cosec(2x)=4$

thanks!

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- Nov 2nd 2010, 01:29 PMBabyMilo3sin(6x)cosec(2x)=4
How would you solve this???

$\displaystyle

3sin(6x)cosec(2x)=4$

thanks! - Nov 2nd 2010, 01:34 PMpickslides
$\displaystyle 3\sin(6x)\cosec(2x)=4$

$\displaystyle \frac{3\sin(6x)}{\sin(2x)}=4$

Do you know the doblue angle formula for sin? - Nov 2nd 2010, 01:39 PMBabyMilo
2sin(x)cos(x)

still dont understand.