Question

Find all solutions to the equation sin^2xcos^2x= (2-sqrt2)/16. Medal and fan

Thanks(:

Answer 1

multiply by 4 both sides

Answer 2

and now think ....

Answer 3

Okay, that's actually what I initially did.

Answer 4

noow we have , (sin 2x)^2 =(2-sqrt2)/4

(sin 2x) =+/-sqrt( (2-sqrt2)/4 )

Answer 5

now try to solve ...

Answer 6

i will do it now ....

Answer 7

Sorry, I was afk, hold on.

Answer 8

Would you divide the right side by 2, right?

Answer 9

ya after taking the sin-1 , you divide by 2

Answer 10

but see also all possible values of 2x in all 4 quadrants

Answer 11

Right, so you would have arcsin(sqrt2 ((2-sqrt2)/4)/2) correct?

Answer 12

u have +/- there see

Answer 13

Yes, I know that. Would that give the correct answers?

Answer 14

ya :) caution : dont miss any of the values of x in any quadrants :)

Answer 15

Right on, thanks!

Answer 16

Here is how I would have done the problem.
It's kind of a weird process, but fun! :)

\[\Large\rm 4\sin^2x \cos^2x=\frac{2-\sqrt2}{4}\]Getting our Double Angle on the left,
and dividing each term by 2 on the right,\[\Large\rm \sin^2\left(2x\right)=\frac{1-\frac{\sqrt2}{2}}{2}\]Taking the square root,\[\Large\rm \sin\left(2x\right)=\pm\sqrt{\frac{1-\frac{\sqrt2}{2}}{2}}\]Relating it to the Half-Angle Formula:\[\Large\rm \sin\left(\frac{\theta}{2}\right)=\pm\sqrt{\frac{1-\cos \theta}{2}}\]If we let our 2x = 4x/2 we can use this identity,\[\Large\rm \sin\left(\frac{4x}{2}\right)=\pm\sqrt{\frac{1-\cos(4x)}{2}}\]Which tells us that:\[\Large\rm \cos(4x)=\frac{\sqrt2}{2}\]Leading to two solution sets for 4x,\[\Large\rm 4x=\frac{\pi}{4}+2k \pi\]\[\Large\rm 4x=\frac{7\pi}{4}+2k \pi\]Dividing by 4 gives us our solutions:\[\Large\rm x=\frac{\pi}{16}+k\frac{\pi}{2}\]\[\Large\rm x=\frac{7\pi}{16}+k\frac{\pi}{2}\]
\(\Large\rm k=0,~\pm1,~\pm2,~...\)

Answer 17

A weird way to get through the problem >.< but tons of fun! hehe

Answer 18

That's what I have!(:

Answer 19

oh cool! c: