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Math Help - I can't solve these Trig Equations

  1. #1
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    Exclamation I can't solve these Trig Equations

    Hi , out of my 100 question holiday packet, these are ones I just couldn't understand:

    Solve these equations in the indicated domain

    (1+cos(theta))/(sin(theta))=-1
    domain: [-180,180)

    cos4(theta) - sin2(theta) = 0
    domain: (-90,90)

    cos4(theta) - sin2(theta) = 1
    domain: [-90,90)

    cos3(theta) + cos5(theta) = 0
    domain: (-90,90)

    sin5(theta) + sin7(theta) = 0
    domain: [-90,90)

    cosx - root3(sinx) = 1
    domain (0,2pi]

    sinx - root3(cosx) = 1
    domain: [-pi,pi]

    [tan(10theta) + tan50(deg)]/[1-tan(10theta)tan50(deg)] = (root3)/3
    domain: (0,90)

    tan(theta) - tan(10theta) = 1 + tan(theta)tan10(deg)
    domain: [-180,180]

    tan(1/2)x + 1 = cosx
    domain: [0,4pi]

    2(cos^2)(1/2)x - 2 = 2cosx
    domain: [-pi,pi)

    2cos(theta + 30)cos(theta - 30) = 1
    domain: [-180,180]

    4sin(theta + 75)cos(theta - 75) = 1
    domain: [-180,180)

    (cos^2)(1/2)x - (1/2)cosx = 1/2
    domain: real numbers

    sinxtan(1/2)x = 1 - cosx
    domain: real numbers

    Any help is greatly appreciated!

    Thanks!
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  2. #2
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    Lexington, MA (USA)
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    Hello, hassapi!

    Here's #8 . . .


    8)\;\;\frac{\tan 10\theta + \tan50^o}{1-(\tan 10\theta)(\tan50^o)} \:=\:\frac{\sqrt{3}}{3}\qquad\text{ domain: }(0^o,90^o)
    We're expected to recognize the left side as a Compound Angle Identity . . .

    . . \tan(10\theta + 50^o) \;=\;\frac{1}{\sqrt{3}}

    And we should know that the angle is 30° and its variations.


    So we have:. . 10\theta + 50^o \;=\;\{30^o,\:210^o,\:390^o,\:570^o,\:750^o,\:930^  o\}

    . . . . . Then: . . . . . 10\theta \;=\;\{{\color{red}\rlap{/////}}-20^o,\:160^o,\:340^o,\;520^o,\:700^o,\:880^o\}

    . . Therefore: . . . . . . \theta \;=\;\{16^o,\;34^o,\:52^o,\:70^o,\:88^o\}

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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, hassapi!

    Here's #8 . . .


    We're expected to recognize the left side as a Compound Angle Identity . . .

    . . \tan(10\theta + 50^o) \;=\;\frac{1}{\sqrt{3}}

    And we should know that the angle is 30° and its variations.


    So we have:. . 10\theta + 50^o \;=\;\{30^o,\:210^o,\:390^o,\:570^o,\:750^o,\:930^  o\}

    . . . . . Then: . . . . . 10\theta \;=\;\{{\color{red}\rlap{/////}}-20^o,\:160^o,\:340^o,\;520^o,\:700^o,\:880^o\}

    . . Therefore: . . . . . . \theta \;=\;\{16^o,\;34^o,\:52^o,\:70^o,\:88^o\}

    Ah, thank you very much! I don't think my teacher taught us compound angle identity...
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