anonymous
  • anonymous
Solve: sin^(2) x - Sin x =0, for 0≤x≤3
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Michele_Laino
  • Michele_Laino
here we have to change variable, namely: \[\Large \sin x = t\]
Michele_Laino
  • Michele_Laino
after that your equation can be rewritten as follows: \[\Large {t^2} - t = 0\]
Michele_Laino
  • Michele_Laino
please solve that equation for t

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
So t=0?
Michele_Laino
  • Michele_Laino
hint, we can factor out t at left side, so we can write: \[\Large t\left( {t - 1} \right) = 0\]
anonymous
  • anonymous
Oh, so t= 0 and 1?
Michele_Laino
  • Michele_Laino
that's right!
Michele_Laino
  • Michele_Laino
now t=0 means: sin x=0 what is x?
anonymous
  • anonymous
x= Radical 3 over 2?
anonymous
  • anonymous
Oops!
Michele_Laino
  • Michele_Laino
no, the solution of that equation: sin x=0, are: \[x = 0,\pi ,2\pi ,3\pi ...\]
Michele_Laino
  • Michele_Laino
nevertheless only x=0, is acceptable, since pi >3
anonymous
  • anonymous
Oh, i get it! Thank you!
Michele_Laino
  • Michele_Laino
please wait, we have to solve this equation: t=1 which means sin x =1
Michele_Laino
  • Michele_Laino
for which values x, we have: sin x= 1?
Michele_Laino
  • Michele_Laino
\[\sin x = 1 \Rightarrow x = ...?\]
anonymous
  • anonymous
Pie over 2 and 3 pie over 2
Michele_Laino
  • Michele_Laino
no, the solutions are: \[x = \frac{\pi }{2},2\pi + \frac{\pi }{2},4\pi + \frac{\pi }{2}...\]
anonymous
  • anonymous
Oh yea, my bad.
Michele_Laino
  • Michele_Laino
nevertheless only x= pi/2 is acceptable
anonymous
  • anonymous
I get it. thanks :D
Michele_Laino
  • Michele_Laino
thanks! :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.