anonymous
  • anonymous
Solve this trigonometric equation. 2sin^2-1=0
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.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Note that this is the same as \[\cos (2x)=0\]
anonymous
  • anonymous
\[\sin(x)=\pm\frac{\sqrt{2}}{2}\] is a start
anonymous
  • anonymous
how didyou get cos(2x)

Looking for something else?

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

More answers

anonymous
  • anonymous
Therefore \[x=\pi/4+n*\pi/2, n \in \mathbb{Z}\]
campbell_st
  • campbell_st
try 2sin^2x = 1 and then sin^2x = 1/2 then go from there
campbell_st
  • campbell_st
so\[\sin(x) = \sqrt{(1/2)} or \pm1/\sqrt{?}\]
campbell_st
  • campbell_st
? = 2
anonymous
  • anonymous
but the restriction id 0 to pi/2
anonymous
  • anonymous
|dw:1327534375645:dw|
campbell_st
  • campbell_st
you should know the angle that gives \[\sin(x)=1/\sqrt{2} or \sin ^{(-1)} (1/\sqrt{2})\]
anonymous
  • anonymous
the angle is 45
anonymous
  • anonymous
thanks i got it ur correct
anonymous
  • anonymous
no problem..happy to help :):)

Looking for something else?

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