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Math Help - solving for all values of x 0° to 360°

  1. #1
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    solving for all values of x 0° to 360°

    Solve for all values of x in the range 0° to 360°

    7sin2x = cosx

    Ive got as far as using a common trig identity of sin2A = 2sincosA

    7(2sinθ cosθ) = cosθ
    14sinθcosθ = cosθ
    14sinθcosθ - cosθ = 0
    cosθ(14sin-1)=0
    either cosθ = 0
    or 14sinθ -1 = 0
    sinθ = 1/14 = 0.071
    cosˆ-1(0) = 90
    θ= 90,270
    sinˆ-1(0.071) = 4.071
    θ= 4.071 and 360-4.071 = 355.929

    Could someone please confirm i have done this the correct way if not please advise.

    Thanks in advance.
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  2. #2
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    Quote Originally Posted by jamm89 View Post
    Solve for all values of x in the range 0° to 360°

    7sin2x = cosx

    Ive got as far as using a common trig identity of sin2A = 2sincosA

    7(2sinθ cosθ) = cosθ
    14sinθcosθ = cosθ
    14sinθcosθ - cosθ = 0
    cosθ(14sin-1)=0
    either cosθ = 0
    or 14sinθ -1 = 0
    sinθ = 1/14 = 0.071
    cosˆ-1(0) = 90
    θ= 90,270
    sinˆ-1(0.071) = 4.071
    θ= 4.071 and 360-4.071 = 355.929

    Could someone please confirm i have done this the correct way if not please advise.

    Thanks in advance.
    I get [tex]sin^{-1}(1/14)= 4.096, not 4.071. Also, sin(360-x)= -sin(x), not sin(x). You want 180- x.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    I get [tex]sin^{-1}(1/14)= 4.096, not 4.071. Also, sin(360-x)= -sin(x), not sin(x). You want 180- x.
    I think the 4.096 not 4.071 was due to rounding but ive changed it.

    So am i right in saying the points that it crosses are 4.096, 90, 175.9 and 180?
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  4. #4
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    Identity problem

    Solve for all values of x in the range 0° to 360° the equation
    7sin2x=cosx

    Could you show me how to do this please.
    I have started by finding the indetity of sin2A = 2sincosA but dont know what to do after that.

    Also Help with this:
    Solve the equation 7cos2θ = 3sinθ for all values of θ in the range 0° to 360°.
    Once again i have selected the trig identity but where to go from there i am struggling.

    Thanks

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  5. #5
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    Hello,
    Quote Originally Posted by jamm89 View Post
    Solve for all values of x in the range 0° to 360° the equation
    7sin2x=cosx

    Could you show me how to do this please.
    I have started by finding the indetity of sin2A = 2sincosA but dont know what to do after that.

    Also Help with this:
    Solve the equation 7cos2θ = 3sinθ for all values of θ in the range 0° to 360°.
    Once again i have selected the trig identity but where to go from there i am struggling.

    Thanks

    So for the first one :
    7sin(2x)=cos(x)
    14sin(x)cos(x)-cos(x)=0
    cos(x)[14sin(x)-1]=0

    and you know that a product ab=0 if and only if either a=0 or b=0 (or both)



    for the second one, use the trig identity for cos(2x) involving only sin(x) :
    cos(2x)=1-2[sin(x)]^2

    so the equation is now :
    7(1-2sin²(x))=3sin(x)
    7-14sin²(x)-3sin(x)=0

    let t=sin(x) and solve the quadratic equation
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  6. #6
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    Quote Originally Posted by jamm89 View Post
    Solve for all values of x in the range 0° to 360° the equation
    7sin2x=cosx

    Could you show me how to do this please.

    I have started by finding the indetity of sin2A = 2sincosA but dont know what to do after that.

    Also Help with this:
    Solve the equation 7cos2θ = 3sinθ for all values of θ in the range 0° to 360°.

    Once again i have selected the trig identity but where to go from there i am struggling.

    [/FONT]
    No need to use so many tags you know

    As you said sin(2x) = 2sin(x)cos(x) so 7sin(2x) = 14sin(x)cos(x) = cos(x)

    You could divide by cos(x) and solve but that won't give you all the solutions, instead take cos(x) from both sides and factorise

    cos(x)(14sin(x)-1) = 0

    then either cos(x) = 0 in which case x = 90, 270 or

    14sin(x)-1 = 0, sin(x) = 1/14 in which case x = 4.10 (3sf)

    so your solutions are 4.10, 90 and 270

    ---------
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  7. #7
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    For the first one i get
    sinΘ= 1/14 = 0.071
    and cosΘ=0 so cosˆ-1 (0)=90

    then sinˆ-1(0.071) = 4.096

    so Θ = 4.096, 90, (180-4.096) and 180

    is that correct?
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  8. #8
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