Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

shaqadry

  • one year ago

solve sin^2 x +2 sinx cosx = 0

  • This Question is Closed
  1. beknazar23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    sinx(sinx+2cosx) = 0 What can you do now? think.

  2. beknazar23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    if a*b = 0, Either a = 0 or b = 0 Here, a = sinx and b = sinx + 2cosx

  3. satellite73
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i don't think this is that easy

  4. satellite73
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    that is \(\sin(x)=0\) is easy enough to solve, you get \(x=n\pi, n\in \mathbb{Z}\) but i can't see a good way to solve \[\sin(x)=-2\cos(x)\]

  5. TheViper
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[sin^2x+2\ sin\ x\ cos\ x=0\] \[sin\ x(sin\ x+2\ cos\ x)=0\] \[(sin\ x+2\ cos\ x)=0\]

  6. satellite73
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    that is not to say you cannot do it, but it is going to take some work

  7. satellite73
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the first thing you have to do to solve \[\sin(x)+2\cos(x)=0\] is to write it as a single function of sine

  8. satellite73
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    in general you can write \[a\sin(x)+b\cos(x)\] as \[\sqrt{a^2+b^2}\sin(x+\theta)\]

  9. tayluvsya
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    :'( i need help on mine

  10. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I kinda wonder if you could do \(\sin^2(x) + \sin(2x)=0\).

  11. satellite73
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    in this case you get \[\sin(x)+2\cos(x)=\sqrt{5}\sin(x+\theta)\] where \(\tan(\theta)=2\)

  12. beknazar23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    tanx = -2 x = atan(-2) + n*(pi)

  13. satellite73
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    therefore \(x+\theta=n\pi\) gives \(\theta=n\pi-\tan^{-1}(2)\)

  14. satellite73
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you can check that it is right by looking here http://www.wolframalpha.com/input/?i=sin%28x%29%2B2cos%28x%29

  15. satellite73
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @ParthKohli i don't think that identity is going to help in this case you will have two arguments, one of \(x\) and the other of \(2x\) if you have a solution from that approach i would love to see it

  16. TheViper
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh I had forgotten this site Thanx @satellite73 for remembering :)

  17. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Gaahh @satellite73 you were right.

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

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

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.