Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

mukushla Group Title

Solve this equation\[\sin^6 x+\cos^6 x=\frac{5}{8}\]for \(x \in[0,2\pi]\)

  • 2 years ago
  • 2 years ago

  • This Question is Closed
  1. sauravshakya Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    sin^6 x + cos^6 x (sin^2x)^3+(cos^2x)^3 (sin^2x + cos^2x) (sin^4x-sin^2x cos^2x +cos^4x) 1*{(sin^2x +cos^2x)^2 -3sin^2x cos^2x} 1-3sin^2xcos^2x 1-3(sinx cosx)^2 1-3(sin2x/2)^2 1-3/4 sin^2(2x)

    • 2 years ago
  2. sauravshakya Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    1-3/4 sin^2 (2x) =5/8 3/4 sin^2 (2x)=3/8 sin^2 (2x)=1/2 2sin^2 (2x) =1 1-2sin^2 (2x)=0 cos4x=0

    • 2 years ago
  3. sauravshakya Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    Now LET a=4x then, cos(a)=0 , 0<=a<=8pi

    • 2 years ago
  4. sauravshakya Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    Solve for a

    • 2 years ago
  5. sauravshakya Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    I hope this will lead to the solution.

    • 2 years ago
  6. sauravshakya Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    And remember x=a/4

    • 2 years ago
  7. sauravshakya Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    Is that correct @mukushla

    • 2 years ago
  8. mukushla Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    nice job :)

    • 2 years ago
  9. harsh314 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    i think if you do it like this it would be shorter \[ (1-\cos^{2}x)^{ 3}\]\[(1-\cos^{2}x)^{ 3} +\cos ^{6}x=\frac{ 5}{ 8 }\] \[1-\cos ^{6}x-3\cos ^{2}x(1-\cos ^{2}x)+\cos ^{6}x=\frac{ 5 }{ 8 }\] the two cos^6 terms cancel \[-3\cos ^{2}x \times \sin ^{2}x=\frac{ 5 }{ 8 }-1\] \[3\sin ^{2}x \times \cos ^{2}x=\frac{ 3 }{ 8 }\] multiply both sides by 4/3 and we have\[4\sin ^{2}xcos ^{2}x=\frac{ 1 }{ 2 }\] and 2sinx cosx=sin2x so \[\sin ^{2}2x=\frac{ 1 }{ 2 }\] hence \[2x=\frac{ \pi }{ 4 } or 2x=\frac{ 3\pi }{ 4 }\] hence x is pi/8 or 3pi/8

    • 2 years ago
  10. harsh314 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    it is looking cumbersome but if you work it out it wont be so.........

    • 2 years ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.