Answers

2014-08-11T17:48:36+05:30
 sin^{4} x+ cos^{4} x = 7 \frac{7sinx.cosx}{2} 

 ( cos^{2} x+ sin^{2}x )^{2} -  ( cos^{2} x- sin^{2} x)^{2} = 7sinx.cosx
 1^{2} + cos^{2} 2x =  \frac{7}{2} ×2sinx.cosx
1+ cos^{2} 2x \frac{7}{2} sin2x.
2- sin^{2} 2x =  \frac{7}{2} sin2x.
2 sin^{2} 2x+7sin2x-4 = 0
after solving this equation,we get,_
sin2x = [tex] \frac{1}{4}  and sin2x = -2
1 5 1
we have sin^4x+cos^4x=(7/2)sinx.cosx
we can write ,2(sin^4x+cos^4x)=(cos^{2}x+sin^{2}x)^2 + (cos^{2}x-sin^{2}x)^2 .bt sorry i wrote (cos^{2}x+sin^{2}x)^2 - (cos^{2}x-sin{2}x)^2 bt next step is right,
ok
thanks
fine