Question

sin^4*x+cos^4*x=7:8 x=??

Answer 1

\[\sin ^{4}x+\cos ^{4}x+2\sin ^{2}x \cos ^{2}x-2\sin ^{2}x \cos ^{2}x=\frac{ 7 }{ 8 }\]
\[\left( \sin ^{2}x+\cos ^{2} x\right)^{2}-2\sin ^{2}x \cos ^{2}x=\frac{ 7 }{8}\]
\[-2\sin ^{2}x \cos ^{2}x=\frac{ 7 }{ 8 }-1=-\frac{ 1 }{8 }\]
\[\left( 2\sin x \cos x \right)^{2}=\frac{ 1 }{ 4 },\sin ^{2}2x=\frac{ 1 }{ 4 },2\sin ^{2}2x=\frac{ 1 }{ 2 }\]
\[1-\cos 4x=\frac{ 1 }{ 2 },\cos 4x=\frac{ 1 }{ 2 }=\cos \frac{ \pi }{3 },\cos \left( \frac{ -\pi}{ 3 } \right)\]
\[4x=2k \pi+\frac{ \pi }{ 3 },2k \pi-\frac{ \pi }{ 3},k=0,1,2,3\]
calculate x

Answer 2

wow man thanks a lot!!

Answer 3

yw