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Solve another trig eqn with square roots

  1. Feb 25, 2005 #1
    solve [tex]\sqrt2 \sin\theta= \sqrt3-\cos\theta [/tex] algebraically for the domain 0<theta<2pi

    I know that the cos can be changed into 1-sin^2 theta but I dont know what to do after how do I get everything on the right hand side and simplify it?
  2. jcsd
  3. Feb 25, 2005 #2
    That cos can't be changed into what you say since its not cos². Anyway, what exactly are you supposed to do with that equation?
  4. Feb 25, 2005 #3
    I need to solve it yes u are right i cant use the trig identity 1+sin^2 theta to replace cos... now im really lost lol what do I do? This is a review question :confused:
  5. Feb 25, 2005 #4


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    If you want to have a cos2 &theta;, then simply manipulate the equation until you get a cos2 &theta;.
  6. Feb 25, 2005 #5
    I dont want to get cos^2 theta i just thought that was the way to solve this but I really have no idea Help plz
  7. Feb 25, 2005 #6


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    The way I prefer to solve this type of problem is to rearrange it to look like:

    A sin x + B cos x = C

    So that I may find a y such that

    A = sin y
    B = cos y

    which allows me to use the addition of angles formula for sine.

    (Note that you need to have A^2 + B^2 = 1 to be able to do this... so you may need to multiply through by a constant so that this will be true)
  8. Feb 27, 2005 #7


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    √2*sin(θ) = √3 - cos(θ)
    ⇒ √2*{1 - cos^2(θ)}^(1/2) = √3 - cos(θ) (<--- sin^2() + cos^2() = 1)
    ⇒ 2*{1 - cos^2(θ)} = 3 - 2*√3*cos(θ) + cos^2(θ) (<--- Squaring Both Sides)
    ⇒ 3*cos^2(θ) - 2*√3*cos(θ) + 1 = 0
    ⇒ {cos(θ) - 1/√3}^2 = 0
    ⇒ θ = arccos(1/√3)
    ⇒ Possible solutions: θ = {(0.9553 rad)=(54.736 deg)} or {(5.328 rad)=(305.264 deg)}
    Checking original equation, only first possibility is solution:
    θ = {(0.9553 rad)=(54.736 deg)}

    Last edited: Feb 27, 2005
  9. Feb 27, 2005 #8
    the solutions are 55 degrees and 305 degrees I found the last post to be a little confusing
  10. Feb 28, 2005 #9
    If you do understand the last post then all you need to do is take the 55° and then minus it from 360° and then you will end up with both answers.

    The Bob (2004 ©)
  11. Mar 2, 2005 #10


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    Where did you become confused? Did you understand the substitution xanthym made for [tex]\sin\theta [/tex] ?

    [tex]\sin\theta = [1 - cos^2\theta]^\frac{1}{2} [/tex] :bugeye:

    With that substitution, the equation became in terms of [tex]\cos\theta [/tex].
    After squaring both sides and rearranging terms, xanthym rewrote the equation so it looks like:
    [tex] ax^2+bx+c = 0 [/tex] ....... [tex] 3cos^2\theta - 2\sqrt{3}\cos\theta + 1 = 0 [/tex]

    Can you factor that equation directly?
    If that looks too tricky, :surprised
    do you know another formula you can use to find roots of an equation in that form?

    Do you understand taking the arccos (inverse cosine) of a value to find the angle?

    (hint: to find both angles, remember there will be more than one quadrant between 0 and 2pi, where cosθ has the same sign). :cool:
  12. Mar 2, 2005 #11


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    xanthym got his one right.That 305° is not a viable solution,bacause it does not satisfy the initial equation.The sine of it is negative,while the the sqrt{3}-\cosine of it is positive...

  13. Mar 3, 2005 #12
    Thanks it took me a little time to absorb but i do understand :approve:
  14. Mar 3, 2005 #13


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    38hrs...That's reasonable...You know what they say:"better late than never".Of course,it does not apply to dating women...

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