anonymous
  • anonymous
How do I solve on the interval from [0, 2pi): 2cos^x-sinx-1=0 ??????? HELP PLEASE
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.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Write it out in latex please :)
anonymous
  • anonymous
\[2\cos ^{2}x-\sin x-1=0\]
saifoo.khan
  • saifoo.khan
Use the identity: \[\sin^2x + \cos^2x =1\]

Looking for something else?

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

More answers

anonymous
  • anonymous
I have been trying that, but I keep getting stuck
saifoo.khan
  • saifoo.khan
2(1-sin^2x) - sinx -1 =0 2 - 2sin^2x - sinx - 1 =0 -2sin^2x - sinx +1 =0 Now suppose sinx as y. Rewrite: -2y^2 - y + 1 = 0 Solve the quadratic.
anonymous
  • anonymous
But I'm solving on an interval of 0 to 2pi, so it needs to be answers like pi/4 pi/6 etc...
saifoo.khan
  • saifoo.khan
Yes. Once you get the value of y, place them in the equation you made: sinx = y. Solve for y.
anonymous
  • anonymous
My teacher never taught us to use y...so now I'm confused out of my mind...
anonymous
  • anonymous
we're working with the unit circle
saifoo.khan
  • saifoo.khan
Alright then. Factor this out: -2sin^2x - sinx +1 =0
anonymous
  • anonymous
would you factor out negative one?
saifoo.khan
  • saifoo.khan
You can do that too. Your answer will be same.
Mertsj
  • Mertsj
\[-2\sin ^2x-\sin x+1=0\]
Mertsj
  • Mertsj
\[2\sin ^2x+\sin x-1=0\]
Mertsj
  • Mertsj
\[(2\sin x-1)(\sin x+1)=0\]
Mertsj
  • Mertsj
Set each factor equal to 0 and solve.

Looking for something else?

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