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Solve for x between 0 , and 2pi the next equation?

  1. Dec 12, 2012 #1
    Solve for x between 0 , and 2pi the next equation?

    cos x + square root(3) sinx = square root(2)

    sin(x+ (pi/6)) = 1/sqrt2 <-- but then I dont know what to do this what I think should do, but Im not sure how to solve x help me thanks!
     
  2. jcsd
  3. Dec 12, 2012 #2

    jedishrfu

    Staff: Mentor

    Re: Trigonometry

    You need to present your homework using the homework template where you state the problem, show some relevant formula that you might use to solve it and show an attempt.

    I don't see how you got the sin(x+pi/6)=1/sqr(2) result from your original equation.
     
  4. Dec 12, 2012 #3

    Mark44

    Staff: Mentor

    Re: Trigonometry

    So sin(x + π/6) = √2/2

    Can you solve sin(u) = √2/2?

    You've done all the hard work. The rest is easy.
     
  5. Dec 12, 2012 #4
    Re: Trigonometry

    Im really confused I dont even know any relevant formula I just tried to move the x
     
  6. Dec 12, 2012 #5
    Re: Trigonometry

    is this right? because somebody said early before that I was wrong... sin (u) = 0.707 ?
     
  7. Dec 12, 2012 #6

    Mark44

    Staff: Mentor

    Re: Trigonometry

    Good advice. luigihs, the template is there for a reason - don't just blow it away...
    He divided both sides of the equation by 2 to get (1/2)sin(x) + (√3/2)cos(x) = √2/2, and then replaced 1/2 by cos(π/6) and √3/2 by sin(π/6). It would have been helpful if he had shown his steps.
     
  8. Dec 12, 2012 #7
    Re: Trigonometry

    Sorry for what is the answer 0.707 ? I dunno how to do the operation
     
  9. Dec 12, 2012 #8

    jedishrfu

    Staff: Mentor

    Re: Trigonometry

    Thanks I didn't see that. I kept thinking to square both sides but then I was still left with a puzzle. Its been a long time since I played with trig problems.
     
  10. Dec 13, 2012 #9

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Re: Trigonometry

    @ luigihs,

    If [itex]\displaystyle \ \sin(\theta)=\frac{\sqrt{2}}{2}\,,\ [/itex] then what is θ ?
     
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