Question

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

Answer 1

change sin in to cos or change cos into sin, and simplify
you will have a quadratic equation in sin square or cos square
solve it, and then find x

Answer 2

i think that might work, but you get a polynomial of degree 4 which will be fairly annoying to solve

Answer 3

maybe try
\[\frac{1}{4}\sin(2x)=\frac{2+\sqrt{2}}{16}\] and go from there. not sure which is easiest

Answer 4

No doubt. Thank goodness you've arrived.

Answer 5

well that was wrong. i mean
\[\frac{1}{4}\sin^2(2x)=\frac{2+\sqrt{2}}{16}\]

Answer 6

still a pain. there is probably some identity that i am missing that would work better. but at least this one gives you
\[\sin^2(2x)=\frac{2+\sqrt{2}}{4}\]

Answer 7

off by a sign
it is
\[\sin^2(2x)=\frac{2-\sqrt{2}}{4}\]

Answer 8

I got one answer as pi/16 and I am not sure if that is correct

Answer 9

we could check it i guess.

Answer 10

dang you are right.
how did you get it?

Answer 11

i am betting experimentx method is actually easiest

Answer 12

i think your method is a lot simpler and better than mine.

Answer 13

well it is still a pain either way

Answer 14

a special case ..!

Answer 15

I got pi/16 because in the beginning I just sqrt both sides and I was left with
sinxcosx=sqrt(2-sqrt(2)/4
and then I just went from there

Answer 16

ok but don't forget \[\sin(x)\cos(x)=-\frac{2-\sqrt{2}}{16}\]

Answer 17

\[(\frac{1-\cos(2x)}{2})(\frac{1+\cos(2x)}{2})=\frac{2=\sqrt{2}}{16}\]

Answer 18

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

Answer 19

\[\sin^2(2x)=\frac{2-\sqrt{2}}{4}\]

Answer 20

But
\[\sin^2\frac{u}{2}=\frac{1-cosu}{2}\]

Answer 21

So
\[\sin ^2(\frac{4x}{2})=\frac{1-\cos (4x)}{2}\]=
\[\frac{2-\sqrt{2}}{4}=\frac{1}{2}-\frac{\frac{\sqrt{2}}{2}}{2}\]
So
\[\cos (4x)=\frac{\sqrt{2}}{2}\]
\[4x=\frac{\pi}{4},\frac{7\pi}{4},\frac{9\pi}{4},\frac{15\pi}{4}, \frac{17\pi}{4},\frac{23\pi}{4},\frac{25\pi}{4}, \frac{31\pi}{4}\]

Answer 22

\[x=\frac{\pi}{16},\frac{7\pi}{16},\frac{9\pi}{16},\frac{15\pi}{16},\frac{17\pi}{16},\frac{23\pi}{16},\frac{25\pi}{16},\frac{31\pi}{16} +2n \pi\]