Question-185257

Question Number 185257 by Kalebwizeman last updated on 19/Jan/23

Answered by SEKRET last updated on 19/Jan/23

\frac{25}{126}

Answered by SEKRET last updated on 19/Jan/23

Q 185134

Commented by Kalebwizeman last updated on 19/Jan/23

o h t h a n k s

Answered by a.lgnaoui last updated on 19/Jan/23

M e t h o d e 1

{(\sin^{2} x + \cos^{2} x)}^{2} = \sin^{4} x + \cos^{4} x + 2 \sin x^{2} \cos x^{2}

1 = \sin^{4} x + \cos^{4} x + \frac{\sin^{2} 2 x}{2} (1)

\cos^{2} x - \sin^{2} x = \cos 2 x = {2 \cos}^{2} x - 1

\cos^{2} x = \frac{\cos 2 x + 1}{2} (2)

(1) \Leftrightarrow 1 = \sin^{4} x + {(\frac{\cos 2 x + 1}{2})}^{2} + \frac{\sin^{2} 2 x}{2}

\frac{{(\cos 2 x + 1)}^{2} + {2 \sin}^{2} 2 x}{4} = 1 - \sin^{4} x (i)

d a p r e s \frac{\sin^{4} x}{5} + \frac{\cos^{4} x}{7} = \frac{1}{12}

\sin^{4} x = \frac{5}{12} - \frac{5 \cos x^{4}}{7}

\frac{5 \cos^{4} x}{7} + \frac{7}{12} = \frac{{(\cos 2 x + 1)}^{2} + {2 \sin}^{2} 2 x}{4}

\frac{{(\cos 2 x + 1)}^{2} + {2 \sin}^{2} 2 x}{4} - \frac{5 {(\cos 2 x + 1)}^{2}}{28} = \frac{7}{12}

[\frac{{(\cos 2 x + 1)}^{2}}{4} - \frac{5 {(\cos 2 x + 1)}^{2}}{28}] + \frac{2 (1 - \cos^{2} 2 x)}{4} = \frac{7}{12}

\cos^{2} 2 x - \frac{\cos 2 x}{3} + \frac{1}{36} = 0

△ = 0

\cos 2 x = \frac{1}{6} \Rightarrow \sin 2 x = \frac{\sqrt{35}}{6}

d o n c

\frac{\sin^{2} 2 x}{5} + \frac{\cos 2 x^{2}}{7} = \frac{\frac{35}{36}}{5} + \frac{\frac{1}{36}}{7}

[\frac{\sin^{2} 2 x}{5} + \frac{\cos^{2} 2 x}{7} = \frac{25}{126}