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Unsure about how to approach this trigonometric question watch

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    Hi All,

    Another question I'm trying to complete, but don't really know how to approach:

    Solve the following for -pi <x pi:

    (1+2cosx)(3tan^2 x - 1) = 0
    I'm not aware of any trigonometric identities I couldn't use to eliminate tan^2, or make 1+2cosx into a form of tan.

    Could someone please assist?
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    (Original post by Ples#!LOL)
    Hi All,

    Another question I'm trying to complete, but don't really know how to approach:

    Solve the following for -pi <x pi:

    (1+2cosx)(3tan^2 x - 1) = 0
    I'm not aware of any trigonometric identities I couldn't use to eliminate tan^2, or make 1+2cosx into a form of tan.

    Could someone please assist?
    You don't need any trig identities for this one. You have two bracketed terms multiplied together to give zero, therefore ...
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    (Original post by Ples#!LOL)
    Hi All,

    Another question I'm trying to complete, but don't really know how to approach:

    Solve the following for -pi <x pi:

    (1+2cosx)(3tan^2 x - 1) = 0
    I'm not aware of any trigonometric identities I couldn't use to eliminate tan^2, or make 1+2cosx into a form of tan.

    Could someone please assist?
    Since the equation is already factorised and equates to zero, you don't need to do any kind of transformation. How would you find x from (x-a)(x-b)=0? You can use the same idea here as a first step.

    Although on the topic of identities for tan, there are some useful ones which you can get by starting from the classic guy:
    \cos^2 x + \sin^2 x =1

    Divide this through by cos^2 x and you get:
    1 + \tan^2 x = \sec^2 x

    Again though, just to reiterate, you don't need the identities here.
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    (Original post by Gregorius)
    You don't need any trig identities for this one. You have two bracketed terms multiplied together to give zero, therefore ...
    Was a bit more busy at work than expected so couldn't get back to solving the problem until now :P

    Would it be cos^-1(- 1/2) = 2/3 pi, so the solutions are 2/3 pi, - 2/3pi
    tan^-1(1/3) = 0.322 radians, pi - 0.322 radians.

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    (Original post by Ples#!LOL)
    Was a bit more busy at work than expected so couldn't get back to solving the problem until now :P

    Would it be cos^-1(- 1/2) = 2/3 pi, so the solutions are 2/3 pi, - 2/3pi
    tan^-1(1/3) = 0.322 radians, pi - 0.322 radians.

    ?
    To find solutions relating to the 2nd bracket, you should be solving for \tan^2 x = \frac{1}{3} (you appear to be solving for \tan x = \frac{1}{3})
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    (Original post by DFranklin)
    To find solutions relating to the 2nd bracket, you should be solving for \tan^2 x = \frac{1}{3} (you appear to be solving for \tan x = \frac{1}{3})
    I see! So it would be tan^2 x = 1/3,
    then take x = tan^-1(√1/3)
    x = 1/6pi

    Then, using the Quadrant rule, 1/6 pi and 5/6 pi are solutions.
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    (Original post by Ples#!LOL)
    I see! So it would be tan^2 x = 1/3,
    then take x = tan^-1(√1/3)
    x = 1/6pi

    Then, using the Quadrant rule, 1/6 pi and 5/6 pi are solutions.
    Half right, but for the wrong reasons.

    Firstly:

    \tan \frac{5\pi}{6}\not= \frac{1}{\sqrt{3}}

    Check which quadrants have tan positive, and what your domain is.

    Secondly:

    As an example.

    If x^2 = 5

    then  x = \pm \sqrt{5}
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    (Original post by ghostwalker)
    Half right, but for the wrong reasons.

    Firstly:

    \tan \frac{5\pi}{6}\not= \frac{1}{3}

    Check which quadrants have tan positive, and what your domain is.

    Secondly:

    As an example.

    If x^2 = 5

    then  x = \pm \sqrt{5}
    Epic fail: I had looked at a different question previously, and thought the range was 0 <= x <= pi :P

    The other solution is - 1/6pi.
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    (Original post by Ples#!LOL)
    Epic fail: I had looked at a different question previously, and thought the range was 0 <= x <= pi :P

    The other solution is - 1/6pi.
    Getting there. -pi/6 is a solution.

    5pi/6 is a solution, but not because tan x =1/root(3), rather -1/root(3)

    There is one other solution you've not mentioned yet. Check the graph of tan over your domain.
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    (Original post by ghostwalker)
    Getting there. -pi/6 is a solution.

    5pi/6 is a solution, but not because tan x =1/root(3), rather -1/root(3)

    There is one other solution you've not mentioned yet. Check the graph of tan over your domain.
    Is -5pi/6 the final solution?
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    (Original post by Ples#!LOL)
    Is -5pi/6 the final solution?
    There is more than one solution
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    (Original post by Desmos)
    There is more than one solution
    Yes, I am aware...

    That was meant as in the final solution that I have not mentioned so far :P
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    (Original post by Ples#!LOL)
    Is -5pi/6 the final solution?
    Yep - that's the missing one.
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    (Original post by Ples#!LOL)
    Yes, I am aware...

    That was meant as in the final solution that I have not mentioned so far :P
    Oh yeah, it is.
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    (Original post by ghostwalker)
    Yep - that's the missing one.
    (Original post by Desmos)
    Oh yeah, it is.
    Thanks guys.
 
 
 
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