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