Question

Find all the solutions to the equation sin^2(x)cos^2(x) = (2 - √2)/16

Answer 1

\[\cos^2(x)\sin^2(x) = 1\]Should I start by transforming: \[\sin^2(x)\cos^2(x)\] into a polynomial? \[\sin^2(x) - \sin^4(x)\]

Answer 2

Thats sin^2(x) + cos^2(x) =1 :P

Answer 3

(1/4)sin^2(2x)=(2-square root 2)/16
sin^2(2x)=(2-square root 2)/4 now by taking roots for both sides
\[\left| \sin(2x) \right| =\frac{ \sqrt{2-\sqrt{2}} }{ 2 }\]
then x=\[x= \pm .5 \sin^{-1}\frac{ \sqrt{2-\sqrt{2}} }{ 2 }\]

Answer 4

oh I forgot the plus!

Answer 5

\(\sin^2(x)\cos^2(x)=(\sin(x)\cos(x))^2\) hence:$$\sin(x)\cos(x)=\pm\frac{\sqrt{2-\sqrt2}}4\\2\sin(x)\cos(x)=\pm\frac{\sqrt{2-\sqrt2}}2\\\sin(2x)=\pm\frac{\sqrt{2-\sqrt2}}2$$hence we get:$$2x=k\pi n+\arcsin\left(\pm\frac{\sqrt{2-\sqrt2}}2\right)=k\pi n\pm\arcsin\left(\frac{\sqrt{2-\sqrt2}}2\right)\\x=\frac{k\pi n}2\pm\frac12\arcsin\left(\frac{\sqrt{2-\sqrt2}}2\right)$$

Answer 6

basically the same approach @Ahmad1 took

Answer 7

I think your way is more simple @oldrin.bataku

Answer 8

Same approach but I noticed that @Ahmad1 had \[x = \pm \frac{ 1 }{ 2 }\sin ^{-1} \frac{ \sqrt{2 - \sqrt{2}} }{ 2 }\] whereas @oldrin.bataku has \[x = \frac{ kπn }{ 2 } \pm \frac{ 1 }{ 2 }\arcsin(\frac{ \sqrt{2-\sqrt{2}} }{ 2 }\]

Answer 9

usually we don't care about the extra term added in @oldrin.bataku ' solution

Answer 10

arcsin means sin^1

Answer 11

OK! Thanks, so it might be easier to just not include it?

Answer 12

well it depends on the interval stated, but in general you can ignore it ;)

Answer 13

Note that:\[\sin^2\left(\frac{\pi}{8}\right)=\frac{1-\cos(\frac{\pi}{4})}{2}=\frac{2-\sqrt{2}}{4}\]\[\Longrightarrow \sin\left(\frac{\pi}{8}\right)=\frac{\sqrt{2-\sqrt{2}}}{2}\]

Answer 14

\[\frac{1}{4}\sin^2(2x)=\frac{2+\sqrt{2}}{16}\]
\[\sin^2(2x)=\frac{2+\sqrt{2}}{4}\]

Answer 15

Oh! That is a simpler way to put it @satellite73 ! Thank you