Math, asked by ggguitarg31, 1 year ago

sin^4(x)+cos^4(x)=7sin(x)cos(x)/2

Answers

Answered by karthik4297

[tex] sin^{4} x+ cos^{4} x = 7 \frac{7sinx.cosx}{2} [/tex]
$( cos^{2} x+ sin^{2}x )^{2} - ( cos^{2} x- sin^{2} x)^{2} = 7sinx.cosx$
$1^{2} + cos^{2} 2x = \frac{7}{2}$ ×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

karthik4297: sorry bro electricity went

karthik4297: in 2nd step sin^4x+cos^4x=(7sinx.cosx)/2,bt i done a mistake i wrote one extra 7

karthik4297: ans is sin2x=1/4 or sin2x=-2

karthik4297: we know that min value of sin2x is -1 so sin2x = -2 is not possible

ggguitarg31: how ( cos^{2} x+ sin^{2}x )^{2} - ( cos^{2} x- sin^{2} x)^{2} = 7sinx.cosx

karthik4297: we have sin^4x+cos^4x=(7/2)sinx.cosx

karthik4297: 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,

ggguitarg31: ok

ggguitarg31: thanks

ggguitarg31: fine

Previous Question

Next Question