Question

sin^4x + cosx^4x = 1 ~.~ someone help please =^= ??

Answer 1

\[\sin^4x + cosx^4x = \sin^4x + (1-\sin^2x)^2\] expand the RHS and then let\[u=\sin^2x\]

And see if you can't get a nice easy little quadratic to solve. :)

Answer 2

uhm.... can't you explain? :< =.=

Answer 3

You are supposed to solve the equation, right?

Answer 4

right ...

Answer 5

Wait - is that a typo? I just realized, you have (and I copied) \[\sin^4x + cosx^4x = 1\] I'm assuming it should be\[\sin^4x + \cos^4x = 1\]

Answer 6

so \[ \cos^4x = (\cos^2x)^2\]
and\[ (\cos^2x)^2=(1-\sin^2x)^2\] right?

Answer 7

yup ! and ... :<

Answer 8

kimmese did u understand the basic concept ?

Answer 9

So we can re-write the equation:\[\sin^4x + (1-\sin^2x)^2 = ?\]
Expand that, and you will have an equation all in sinx

Answer 10

I should say, "we can rewrite the LHS of the equation...."

Answer 11

uhm cos^2x + sin^2 x =1 right ? so sin^2x = 1 - cos^2x right ... :<

Answer 12

Yes, sorry. That's the fundamental ID, I was assuming that didn't need to be stated. :)

Answer 13

:> =.= now i got it =.= ! thanks =.= !!

Answer 14

you're welcome :)

Answer 15

I'm kind of fond of a slightly different method.

Take \(\sin^{2}(x) + \cos^{2}(x) = 1\)

Everything is positive, so there is no harm in squaring.

\(\sin^{4}(x) + 2\sin^{2}(x)\cos^{2}(x) + \cos^{4}(x) = 1\)

From the original equation, this gives:

\(2\sin^{2}(x)\cos^{2}(x) = 0\)

Answer 16

thank you =)