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Math Help - Trig Problem

  1. #1
    Newbie Tom G's Avatar
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    Trig Problem

    There are two similar questions in my homework which need me to solve equations and I have now got stuck on one of them and would like someone to tell me if I have made a mistake in my workings or what I should do next.

    Q. Solve 2tan^2 x-7sec+8=0 for 0<x<360

    So far I have got:

    2tan^2 x-7sec+8=0

    2(sin^2 x/cos^2 x)-7cosx+8=0

    I am mainly confused because I want to multiply by cos^2 x in order to cancel out the cos^2 x under sin^2 x, but can I do this given that it is all multiplied by 2 i.e. in the brackets?


    ALSO: I would be grateful if someone could clarify this for me.

    Q. State the values of:

    arc sin 0.5 - my answer is 30 degrees

    arc tan 1 - my answer is 45 degrees

    I ask this because I have no idea what the 'arc' means - does it affect the way I should answer the question as I have never encountered it before.

    Many thanks in advance MHF!
    Last edited by Tom G; September 20th 2007 at 11:21 AM. Reason: correct error in post
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  2. #2
    MHF Contributor
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    2tan^2 x-7sec+8=0

    2(sin^2 x/cos^2 x)-7cosx+8=0

    That's okay except that it should be -7/cos x.

    You wanted to multiply all by cos^2 x in order to clear the fraction? Very good! Always clear the fractions first.

    Sure you can multiply all by cos^2 x even if "it is all multiplied by 2 i.e. in the brackets".

    So, to continue,

    2[sin^2(X) /cos^2(X)] -7/cosX +8 = 0
    Multiply both sides by cos^2(X),
    2sin^2(X) -7cosX +8cos^2(X) = 0
    2[1 -cos^2(X)] -7cosX +8cos^2(X) = 0
    2 -2cos^2(X) -7cosX +8cos^2(X) = 0
    6cos^2(X) -7cosX +2 = 0
    Factor that,
    (3cosX -2)(2cosX -1) = 0

    3cosX -2 = 0
    cosX = 2/3
    X = arccos(2/3) = 48.1896851 degrees, in the 1st quadrant.
    Since cosine is positive also in the 4th quadrant,
    X = 360 -48.1896851 = 311.8103149 deg, in the 4th quadrant.

    2cosX -1 = 0
    cosX = 1/2
    X = arccos(1/2)) = 60, in the 1st quadrant.
    Since cosine is positive also in the 4th quadrant,
    X = 360 -60 = 300 deg, in the 4th quadrant.

    Therefore, X = 48.1896851, 60, 300, or 311.8103149 degrees -----answer.

    -----------------------------
    ALSO: I would be grateful if someone could clarify this for me.

    Q. State the values of:

    arc sin 0.5 - my answer is 30 degrees

    arc tan 1 - my answer is 45 degrees

    I ask this because I have no idea what the 'arc' means - does it affect the way I should answer the question as I have never encountered it before.


    arcsin(0.5) = 30 deg or 150 deg, since sine is positive in the 1st and 2nd quadrants.

    arctan(1) = 45deg or 225deg, since tangent is positive in the 1st and 3rd quadrants.

    'arc' here means, or, arcsin(0.5) means an anlge whose sine is 0.5.
    arctan(1) is an angle whose tangent is 1.
    So associate "arc___" with an angle.
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  3. #3
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    Hello, Tom G!

    Here's the first one . . .


    1) Solve: . 2\tan^2 x -7\sec x + 8\;=\;0 . for 0^o \leq x \leq 360^o

    We have: . 2\left(\sec^2x - 1\right) - 7\sec x + 8 \;=\;0

    . . which simplifies to: . 2\sec^2x - 7\sec x + 6 \;=\;0

    . . which factors: . (\sec x - 2)(2\sec x - 3) \;=\;0


    And has roots:

    . . \sec x - 2 \:=\:0\quad\Rightarrow\quad\sec x \:=\:2\quad\Rightarrow\quad x \:=\:60^o,\:300^o

    . . 2\sec x - 3 \:=\:0\quad\Rightarrow\quad\sec x \:=\:\frac{3}{2}\quad\Rightarrow\quad x \:\approx\:48.19^o,\:311.81^o

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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Tom G View Post
    Q. State the values of:

    arc sin 0.5 - my answer is 30 degrees

    arc tan 1 - my answer is 45 degrees

    I ask this because I have no idea what the 'arc' means - does it affect the way I should answer the question as I have never encountered it before.
    The "arcsine" function is the "older" labeling of the inverse sine function. ie.
    sin(30^o) = \frac{1}{2} \implies arcsin \left ( \frac{1}{2} \right ) = sin^{-1} \left ( \frac{1}{2} \right) = 30^o

    You won't typically see the newer textbooks using asin, acs, atn, etc. (Please don't ask me for a cut-off date for this. For all I know there are some new texts using this.) For example I learned the "arc" functions in High School in the late 80's, but my Calc book in college didn't use them. However my Calc book that was published in the 70's does use them.

    -Dan
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  5. #5
    Newbie Tom G's Avatar
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    Just one more question...

    Does arc cos(-1/square root of 2) = -45 ?
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Tom G View Post
    Just one more question...

    Does arc cos(-1/square root of 2) = -45 ?
    no. (remember cosine is an even function)

    couldn't you have plugged that in to your calculator? or you could work it out (it's actually not that hard)
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  7. #7
    MHF Contributor
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    Quote Originally Posted by Tom G View Post
    Just one more question...

    Does arc cos(-1/square root of 2) = -45 ?
    -45 degrees?
    No.

    -1/sqrt(2) is negative. ()
    In what quadrants is cosine negative?
    In the 2nd and 3rd quadrants.
    So, 135 deg and 225 deg. --------------answer.
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