- #1
mirandab17
- 40
- 0
Solve: sin^2x = (sinx)(cosx)
0 < x < 2pi
Find exact solutions.
Okay, so I got 0, pi/4, but just wasn't sure how to do this...
I know you bring the sinxcosx to the other side, making it
sin^2x - sinxcosx = 0
...and then I tried factoring out sinx leaving me with sinx(sinx - cosx). That's where I got lost. So sinx is 0 at 0 and pi on the unit circle, so two solutions are 0 and pi, but what on Earth do I do with sinx - cosx? That's when I started thinking about possible other identities that could help me out, but each left me with more craziness on my page!
0 < x < 2pi
Find exact solutions.
Okay, so I got 0, pi/4, but just wasn't sure how to do this...
I know you bring the sinxcosx to the other side, making it
sin^2x - sinxcosx = 0
...and then I tried factoring out sinx leaving me with sinx(sinx - cosx). That's where I got lost. So sinx is 0 at 0 and pi on the unit circle, so two solutions are 0 and pi, but what on Earth do I do with sinx - cosx? That's when I started thinking about possible other identities that could help me out, but each left me with more craziness on my page!