Question

Solve this equation\[\sin^6 x+\cos^6 x=\frac{5}{8}\]for \(x \in[0,2\pi]\)

Answer 1

sin^6 x + cos^6 x
(sin^2x)^3+(cos^2x)^3
(sin^2x + cos^2x) (sin^4x-sin^2x cos^2x +cos^4x)
1*{(sin^2x +cos^2x)^2 -3sin^2x cos^2x}
1-3sin^2xcos^2x
1-3(sinx cosx)^2
1-3(sin2x/2)^2
1-3/4 sin^2(2x)

Answer 2

1-3/4 sin^2 (2x) =5/8
3/4 sin^2 (2x)=3/8
sin^2 (2x)=1/2
2sin^2 (2x) =1
1-2sin^2 (2x)=0
cos4x=0

Answer 3

Now LET a=4x then,
cos(a)=0 , 0<=a<=8pi

Answer 4

Solve for a

Answer 5

I hope this will lead to the solution.

Answer 6

And remember x=a/4

Answer 7

Is that correct @mukushla

Answer 8

nice job :)

Answer 9

i think if you do it like this it would be shorter
\[ (1-\cos^{2}x)^{ 3}\]\[(1-\cos^{2}x)^{ 3} +\cos ^{6}x=\frac{ 5}{ 8 }\]
\[1-\cos ^{6}x-3\cos ^{2}x(1-\cos ^{2}x)+\cos ^{6}x=\frac{ 5 }{ 8 }\]
the two cos^6 terms cancel
\[-3\cos ^{2}x \times \sin ^{2}x=\frac{ 5 }{ 8 }-1\]
\[3\sin ^{2}x \times \cos ^{2}x=\frac{ 3 }{ 8 }\]
multiply both sides by 4/3
and we have\[4\sin ^{2}xcos ^{2}x=\frac{ 1 }{ 2 }\]
and 2sinx cosx=sin2x
so \[\sin ^{2}2x=\frac{ 1 }{ 2 }\]
hence \[2x=\frac{ \pi }{ 4 } or 2x=\frac{ 3\pi }{ 4 }\]
hence x is pi/8 or 3pi/8

Answer 10

it is looking cumbersome but if you work it out it wont be so.........