Results 1 to 5 of 5

Math Help - Weird MCQ options!

  1. #1
    Super Member
    Joined
    Dec 2009
    Posts
    755

    Weird MCQ options!

    Determine all the values of angle x if  3sinx=tanx and  x is between  0^\circ and  360^\circ. Give the answers correct to 1 decimal place.

    [A] 70.5^\circ, 180^\circ, 250.5^\circ

    [B] 0^\circ, 70.5^\circ, 180^\circ, 289.5^\circ, 360^\circ

     [C] 0^\circ, 70.5^\circ, 180^\circ, 250.5^\circ, 360^\circ

     [D] 70.5^\circ, 180^\circ, 289.5^\circ

    This is weird, I have solved to the point that cosx=\frac{1}{3} and  x is in Q_1, Q_4. Hence one of the options should have only 2 answers. Did I go wrong?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by Punch View Post
    Determine all the values of angle x if  3sinx=tanx and  x is between  0^\circ and  360^\circ. Give the answers correct to 1 decimal place.

    [A] 70.5^\circ, 180^\circ, 250.5^\circ

    [B] 0^\circ, 70.5^\circ, 180^\circ, 289.5^\circ, 360^\circ

     [C] 0^\circ, 70.5^\circ, 180^\circ, 250.5^\circ, 360^\circ

     [D] 70.5^\circ, 180^\circ, 289.5^\circ

    This is weird, I have solved to the point that cosx=\frac{1}{3} and  x is in Q_1, Q_4. Hence one of the options should have only 2 answers. Did I go wrong?
    You didn't go wrong but rather didn't complete the whole equation. The problem you made appears to be that you cancelled out sin(x) which you can't do because sin(x) is potentially 0. I've outlined the method using factorising below

    From the domain restrictions x \neq 90^{\circ} + 180k \: \: , \: k \in \mathbb{Z}

    3\sin(x) - \frac{\sin(x)}{\cos(x)} = 0

    \frac{3\sin(x) \cos(x) - \sin(x)}{cos(x)}

    Multiply by cos(x) and factorise

    \sin(x) (3\cos(x)-1) = 0

    \sin(x) = 0 or \cos(x) = \frac{1}{3}

    x = 0\: ,\: 180\: ,\: 360

    x = 70.5\: , \: 289.5

    This is because cos(x) is symmetrical in the line x = 180^{\circ} so we can do 180+(180-70.5) where 180-70.5 is the distance the first point is away from 180
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Dec 2009
    Posts
    755
    Quote Originally Posted by e^(i*pi) View Post
    You didn't go wrong but rather didn't complete the whole equation. The problem you made appears to be that you cancelled out sin(x) which you can't do because sin(x) is potentially 0. I've outlined the method using factorising below

    From the domain restrictions x \neq 90^{\circ} + 180k \: \: , \: k \in \mathbb{Z}

    3\sin(x) - \frac{\sin(x)}{\cos(x)} = 0

    \frac{3\sin(x) \cos(x) - \sin(x)}{cos(x)}

    Multiply by cos(x) and factorise

    \sin(x) (3\cos(x)-1) = 0

    \sin(x) = 0 or \cos(x) = \frac{1}{3}

    x = 0\: ,\: 180\: ,\: 360

    x = 70.5\: , \: 289.5

    This is because cos(x) is symmetrical in the line x = 180^{\circ} so we can do 180+(180-70.5) where 180-70.5 is the distance the first point is away from 180
    So i Cant cancel sinx
    Follow Math Help Forum on Facebook and Google+

  4. #4
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by Punch View Post
    So i Cant cancel sinx
    Nope. Attached is the graph of this question, on it you can see that 0 and 180 are definitely crossing points.

    If you'd cancelled sin(x) you'd have been dividing by 0 at the points 0, 180 and 360.

    In general it's always better to look for a factorising solution first
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Dec 2009
    Posts
    755
    I will move everything to the left side or right side before factorizing the common items and equating them to values before solving... thanks!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Looking for options equal to the equation
    Posted in the Algebra Forum
    Replies: 9
    Last Post: November 13th 2011, 04:57 AM
  2. confused by all the options
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: January 30th 2010, 05:14 AM
  3. Which options gives f(7-x)
    Posted in the Algebra Forum
    Replies: 3
    Last Post: December 27th 2009, 03:33 PM
  4. Call Options and Strikes
    Posted in the Business Math Forum
    Replies: 0
    Last Post: May 26th 2008, 06:46 AM

Search Tags


/mathhelpforum @mathhelpforum