i need some help resolvong $\displaystyle \cos^2 x \sin^2 x = 0$ over the interval [0 degrees and 360 degrees]

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- Apr 14th 2009, 06:22 PMdarkmoon123help: solving equation (cos^2x)(sin^2x)=0
i need some help resolvong $\displaystyle \cos^2 x \sin^2 x = 0$ over the interval [0 degrees and 360 degrees]

- Apr 14th 2009, 06:49 PMmr fantastic
- Apr 14th 2009, 06:49 PMdarkmoon123
I think i have the answer but i am not sure, can someone check this:

$\displaystyle (cos^2x)(sin^2x)=0$

$\displaystyle (1-sin^2x ) = 0 $ $\displaystyle (1-cos^2x) = 0$

$\displaystyle sin^2x=1$ square root both sides $\displaystyle cos^2x=1 $ square root both sides

$\displaystyle sinx=1 $ $\displaystyle cosx=1 $

{90 degrees, 0 degrees} ? - Apr 14th 2009, 06:54 PMmr fantastic
- Apr 15th 2009, 05:40 AMstapel
If you want to use this method, instead of the one explained earlier, you'll need to use what you learned back in algebra about

**solving quadratic equations**.

In particular, if $\displaystyle 1\, -\, \sin^2(x)\, =\, 0$, then $\displaystyle (1\, -\, \sin(x))(1\, +\, \sin(x))\, =\, 0$, so $\displaystyle \sin(x)\, =\, \pm 1$. And so forth.

(Wink)