1. ## analytic trig

yeah i dont understand. why there are no solutions?

greater than 0??

2. Originally Posted by algebra2

yeah i dont understand. why there are no solutions?

greater than 0??
do you realize that $\sin^2 x$ is always non-negative? it's a square term, so it is greater than or equal to zero (it is also less than or equal to 1, to be precise)

$2 \sin^2 x + 1 = 0$

means $2 \sin^2 x = -1$

but -1 is negative, thus we cannot solve this for any real number x

got it?

3. yea thanks. i kind of overlooked that i guess.