Page 1 of 2 12 LastLast
Results 1 to 10 of 11

Thread: Trig equation: cos(3π/2−2x)+sin(π−3x)=sin(3x)−cos(x)

  1. #1

    Trig equation: cos(3π/2−2x)+sin(π−3x)=sin(3x)−cos(x)

    You can see question in photo below

    Please tell me identities and/or other stuff to solve this.
    20170619_191130.jpg

  2. #2
    Full Member
    Join Date
    Mar 2016
    Posts
    839
    Quote Originally Posted by Tuharramah View Post
    You can see question in photo below

    Please tell me identities and/or other stuff to solve this.
    Okay, just so we'll all on the same page, this is what I think your image says the problem is. If it's incorrect, please reply with any necessary corrections:

    [tex]cos \left( \dfrac{3\pi}{2}-2x \right) + sin \left( \pi - 3x \right) = sin(3x) - cos(x)[/tex]

    I'll assume that you've read the Read Before Posting thread, and also assume that because you've shown no work of your own, you have none to share with us and need help at the very beginning. I'd begin by revisiting your class notes and/or textbook to find any identities you've been taught in the class so far. In particular, I'd look for the sum and difference identities. They should look something like: [tex]sin(\alpha + \beta) = \text{...}[/tex]. If, for whatever reason, you can't find them in your notes or textbook, you can try this webpage for a refresher.

    Once you've reviewed these identities, go ahead and give the problem your best effort. Then please include any and all work you've done on this problem, even the parts you know for sure are wrong. Thank you.

  3. #3
    Elite Member
    Join Date
    Jun 2007
    Posts
    17,066
    Quote Originally Posted by Tuharramah View Post
    You can see question in photo below

    Please tell me identities and/or other stuff to solve this.
    20170619_191130.jpg
    cos(3π/22x)+sin(π3x)=sin(3x)cos(x)

    Useful identities for this problem:

    cos(π/2 - Θ) = ?

    cos(π - Θ) = ?

    sin(π - Θ) = ?

    sin(3Θ) = ? [as a function of sin(Θ) and cos(Θ)]

    cos(2Θ) = ? [as a function of sin(Θ) and cos(Θ)]
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  4. #4

    By studying your suggestion

    I got x as 30 deg.
    Unfortunately i couldn't upload my solution
    So is it correct answer?

  5. #5
    Elite Member
    Join Date
    Jun 2007
    Posts
    17,066
    Quote Originally Posted by Tuharramah View Post
    I got x as 30 deg.
    Unfortunately i couldn't upload my solution
    So is it correct answer?
    If you put x = 30°, do you satisfy the given equation:
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  6. #6

    No :(

    < link to objectionable site removed >

    Here is my solution
    As I said I couldn't upload it here
    Last edited by mmm4444bot; 09-02-2017 at 06:55 PM.

  7. #7
    Full Member
    Join Date
    Mar 2016
    Posts
    839
    Your solution of [tex]x = 30^{\circ}=\dfrac{\pi}{6}[/tex] is a valid solution...

    [tex]cos \left( \dfrac{3\pi}{2} - \dfrac{2 \pi}{6} \right) + sin \left( \pi - \dfrac{3 \pi}{6} \right) = sin\left( \dfrac{3 \pi}{6} \right) - cos\left( \dfrac{\pi}{6} \right)[/tex]

    [tex]cos \left( \dfrac{7 \pi}{6} \right) + sin \left( \dfrac{\pi}{2} \right) = sin\left( \dfrac{3 \pi}{6} \right) - cos\left( \dfrac{\pi}{6} \right)[/tex]

    [tex]-\dfrac{\sqrt{3}}{2} + 1 = 1 - \dfrac{\sqrt{3}}{2}[/tex]

    [tex]0.134 \approx 0.134[/tex]

    ...but it's not the only one. I'm assuming you're only interested in solutions for [tex]0 \le x \le 2\pi[/tex] or [tex]0^{\circ} \le x \le 360^{\circ}[/tex]. There's another solution to the equation [tex]sin(x)=\dfrac{1}{2}[/tex] on that interval. Plus, there's two other solutions this equation won't find you.

    Your work is fine up until you get to this point:

    [tex]-2sin(x)cos(x) + cos(x) = 0[/tex]

    From here, what happens if you multiply by -1 and factor?

    [tex]cos(x) \cdot (2sin(x) - 1) = 0[/tex]

    Where does that lead you?

  8. #8
    So it will be:
    -2sin (x)×cos (x)+cos (x)=0

    2sin (x)×cos (x)-cos (x)=0

    Cos (x) × (2sin (x) -1)=0

    X1=90
    X2=30
    ..
    So when I divided by cos x ,I killed one solution

  9. #9
    Elite Member
    Join Date
    Jun 2007
    Posts
    17,066
    Quote Originally Posted by Tuharramah View Post
    So it will be:
    -2sin (x)×cos (x)+cos (x)=0

    2sin (x)×cos (x)-cos (x)=0

    Cos (x) × (2sin (x) -1)=0

    X1=90
    X2=30
    ..
    So when I divided by cos x ,I killed one solution
    Was your domain restricted like: 0 ≤ x ≤ π/2
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  10. #10
    Let's suppose it is.
    Because no information about domain is given

Bookmarks

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •