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Thread: Trig equations

  1. #1
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    Trig equations

    Hello,

    I am having a bit of trouble solving this eqn:

    tan (x) + cot (x) = 2sec (x)
    For all angles between 0 and 360 degrees.

    Thank you!

    Max
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  2. #2
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    Quote Originally Posted by darksupernova View Post
    Hello,

    I am having a bit of trouble solving this eqn:

    tan (x) + cot (x) = 2sec (x)
    For all angles between 0 and 360 degrees.

    Thank you!

    Max
    Dear darksupernova,

    $\displaystyle tanx+cotx=2secx$

    $\displaystyle \frac{1}{sin(x)cos(x)}=2secx$

    $\displaystyle 2sinx=1$

    $\displaystyle sinx=\frac{1}{2}$

    Since x is inbetween 0 and 360 degrees,

    $\displaystyle x=30^0$

    Hope this will help you.
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  3. #3
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    Ah very clever!

    Thank you very much!

    Im now a but stuck on this:

    tan (x) + 3cot (x) = 5sec (x)

    Very similar but the cos^2 {x) + sin^2 (X) = 1 trick doesnt work?
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  4. #4
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    Quote Originally Posted by darksupernova View Post
    Hello,

    I am having a bit of trouble solving this eqn:

    tan (x) + cot (x) = 2sec (x)
    For all angles between 0 and 360 degrees.

    Thank you!

    Max
    $\displaystyle tanx+cotx=\frac{sinx}{cosx}+\frac{cosx}{sinx}=\fra c{sinx}{sinx}\ \frac{sinx}{cosx}+\frac{cosx}{cosx}\ \frac{cosx}{sinx}$

    $\displaystyle =\frac{sin^2x+cos^2x}{sinxcosx}$

    $\displaystyle \Rightarrow\ \frac{1}{sinxcosx}=\frac{2}{cosx}\ \Rightarrow\ \frac{1}{sinx}=2\ \Rightarrow\ sinx=\frac{1}{2}$

    The vertical co-ordinate of the angle is 0.5 in the unit circle, centre (0,0).
    This corresponds to two angles,

    $\displaystyle x=sin^{-1}\left(\frac{1}{2}\right),\ x={\pi}-sin^{-1}\left(\frac{1}{2}\right)$

    $\displaystyle x=30^o,\ x=150^o$
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  5. #5
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    Quote Originally Posted by darksupernova View Post
    Ah very clever!

    Thank you very much!

    Im now a but stuck on this:

    tan (x) + 3cot (x) = 5sec (x)

    Very similar but the cos^2 {x) + sin^2 (X) = 1 trick doesnt work?
    $\displaystyle \frac{sinx}{cosx}+\frac{3cosx}{sinx}=\frac{5}{cosx }$

    $\displaystyle \frac{sin^2x+3cos^2x}{cosxsinx}=\frac{5}{cosx}$

    $\displaystyle \frac{sin^2x+cos^2x+2cos^2x}{cosxsinx}=\frac{1+2co s^2x}{cosxsinx}=\frac{5}{cosx}$

    $\displaystyle \frac{1+2cos^2x}{sinx}=5$

    $\displaystyle 2cos^2x+1=5sinx$

    $\displaystyle 2-2sin^2x-5sinx+1=0$

    $\displaystyle 2sin^2x+5sinx-1=0$

    $\displaystyle sinx=\frac{-5\pm\sqrt{25+8}}{4}$

    x is the inverse sine of the 2 results.
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  6. #6
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    Another clever little trick on line three, hadnt thought of that, thankyou very much! Now that ive seen these tricks i wont forget them!
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