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Math Help - trigonometric inequality

  1. #1
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    trigonometric inequality

    Find the value of x , valid between -pi and pi , for the following inequality . (I am not allowed to do these using the graphical method)

    (1) tan x < cos x


    (2) 2 cos 2x > 3cos x -1

    I just do not know the steps .The range of x makes it worse . THanks for your time .
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  2. #2
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    Quote Originally Posted by thereddevils View Post
    Find the value of x , valid between -pi and pi , for the following inequality . (I am not allowed to do these using the graphical method)

    (1) tan x < cos x


    (2) 2 cos 2x > 3cos x -1

    I just do not know the steps .The range of x makes it worse . THanks for your time .
    (1) Consider:

    \tan x = \cos x \Rightarrow \frac{\sin x}{\cos x} = \cos x

     \Rightarrow \sin x = \cos^2 x \Rightarrow \sin x = 1 - \sin^2 x

     \Rightarrow \sin^2 x + \sin x - 1 = 0

    \Rightarrow \sin x = \frac{-1 + \sqrt{5}}{2} (the negative root solution is rejected - why?)

    \Rightarrow x = \theta_1 = \sin^{-1} \left( \frac{-1 + \sqrt{5}}{2}\right) or x = \theta_2 = \pi - \sin^{-1} \left( \frac{-1 + \sqrt{5}}{2}\right).

    This breaks the interval [-\pi, \pi] into the several subintervals. Pick a simple value in each of the subintervals and use it to test whether the inequality is true or not for that subinterval.


    (2) Done in a similar way.

    Note: 2 \cos (2x) = 3 \cos x - 1 \Rightarrow 2 \cos^2 x - 1 = 3 \cos x - 1 \Rightarrow 2 \cos^2 x - 3 \cos x = 0.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    (1) Consider:

    \tan x = \cos x \Rightarrow \frac{\sin x}{\cos x} = \cos x

     \Rightarrow \sin x = \cos^2 x \Rightarrow \sin x = 1 - \sin^2 x

     \Rightarrow \sin^2 x + \sin x - 1 = 0

    \Rightarrow \sin x = \frac{-1 + \sqrt{5}}{2} (the negative root solution is rejected - why?)

    \Rightarrow x = \theta_1 = \sin^{-1} \left( \frac{-1 + \sqrt{5}}{2}\right) or x = \theta_2 = \pi - \sin^{-1} \left( \frac{-1 + \sqrt{5}}{2}\right).

    This breaks the interval [-\pi, \pi] into the several subintervals. Pick a simple value in each of the subintervals and use it to test whether the inequality is true or not for that subinterval.


    (2) Done in a similar way.

    Note: 2 \cos (2x) = 3 \cos x - 1 \Rightarrow 2 \cos^2 x - 1 = 3 \cos x - 1 \Rightarrow 2 \cos^2 x - 3 \cos x = 0.

    Thanks Mr F . .

    For (1) , i found x1 and x2 to be 38.17 and 141.83 respectively .

    Then i draw a line with 4 'points', [ -pi , 38.17 , 141.83 , pi ] and tested them

    i got (-pi,38.17) as the solution but this doesn't agree with my answer .

    Did i miss out any sub intervals ? Thanks .
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  4. #4
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    Quote Originally Posted by thereddevils View Post
    Thanks Mr F . .

    For (1) , i found x1 and x2 to be 38.17 and 141.83 respectively . Mr F says: These answers are in degrees. You're meant to get them in radians (see my reply in one of your other threads).

    Then i draw a line with 4 'points', [ -pi , 38.17 , 141.83 , pi ] and tested them

    i got (-pi,38.17) as the solution but this doesn't agree with my answer .

    Did i miss out any sub intervals ? Thanks .
    The solution will be the following intervals:

    -\pi \leq x < -\frac{\pi}{2}, 0.66624 < x < \frac{\pi}{2} and 2.47535 < x \leq \pi

    (correct to 5 decimal places).

    Drawing some graphs really does help show what's going on ....
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    The solution will be the following intervals:

    -\pi \leq x < -\frac{\pi}{2}, 0.66624 < x < \frac{\pi}{2} and 2.47535 < x \leq \pi

    (correct to 5 decimal places).

    Drawing some graphs really does help show what's going on ....
    Well thanks Mr F .. be patient with me ..

    I think my problem is not at solving the inequality or drawing the graph .It's at solving the equation itself to find the intersection of the graphs , especially when its range is from -pi to pi . I don seem to understand how to find the x's when it goes clockwise (-ve)

    where did the pi/2 come from ?
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  6. #6
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    Quote Originally Posted by thereddevils View Post
    Well thanks Mr F .. be patient with me ..

    I think my problem is not at solving the inequality or drawing the graph .It's at solving the equation itself to find the intersection of the graphs , especially when its range is from -pi to pi . I don seem to understand how to find the x's when it goes clockwise (-ve)

    where did the pi/2 come from ?
    Draw the graphs of y = tan x and y = cos x. Note that tan x is undefined for x = pi/2, -pi/2 etc. ....
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