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Trig identity and equations

  1. Apr 17, 2005 #1
    how can i do this trig identity

    ln|sec(theta) + tan(theta| + ln|sec(theta) + tan(theta)| = 0

    and this trig equation
    4sin(squared)x + 2cos(squared)x = 3

    and cos2(theta) + 3 = 5cos(theta)


    any help will be appreciated
     
  2. jcsd
  3. Apr 17, 2005 #2

    dextercioby

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    Rewrite the first.It doesn't make any sense.

    For the second,use [itex] \cos^{2}x=1-\sin^{2}x [/itex] for transform it into a quadratic in "sin x".

    For the third,use [itex] \cos 2x=2\cos^{2}x-1 [/tex] to transform it into a quadratic in "cos x".

    Daniel.
     
  4. Apr 17, 2005 #3
    ln|sec(theta) + tan(theta)| + ln|sec(theta) - tan(theta)| = 0

    | (line) is absolute value
     
  5. Apr 17, 2005 #4

    dextercioby

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    There's no big deal

    [tex] \ln\left|\sec\theta+\tan\theta\right|+\ln\left|\sec\theta-\tan\theta\right|=\ln|\sec^{2}\theta-\tan^{2}\theta|=\ln 1 =0 [/tex]

    Q.e.d.

    Daniel.
     
  6. Apr 18, 2005 #5

    uart

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    Every one of those equations is nonsense. Sure you need help, you need help to transcribe your questions accurately. You are doomed to failure in any problem if you cant even write the problem description properly.

    BTW. Congrats to dextercioby for figuring out what you actually meant to write in at least one of those problems.
     
  7. Apr 18, 2005 #6
    Practical Practice...


    Trig identity confirmed:
    [tex]\boxed{\ln\left|\sec\theta+\tan\theta\right|+\ln\left|\sec\theta-\tan\theta\right|=\ln|\sec^{2}\theta-\tan^{2}\theta|=\ln 1 =0} [/tex]

    [tex] 4 \sin^2 x + 2 \cos^2 x = 3 [/tex]
    [tex] \cos^2 x = 1 - \sin^2 x [/tex] - identity
    [tex] 4 \sin^2 x + 2(1 - \sin^2 x) - 3 = 0 [/tex]
    [tex] 4 \sin^2 x + 2 - 2 \sin^2 x - 3 = 0 [/tex]
    [tex] 4 \sin^2 x - 2 \sin^2 x - 1 = 0 [/tex]
    [tex] 2 \sin^2 x - 1 = 0 [/tex]
    [tex] \sin^2 x = \frac{1}{2} [/tex]
    [tex] \boxed{\sin x = \pm \frac{\sqrt{2}}{2}} [/tex]
    :biggrin:

    [tex] \cos 2x + 3 = 5 \cos x [/tex]
    [tex] \cos 2x - 5 \cos x + 3 = 0 [/tex]
    [tex] \cos 2x = 2 \cos^2 x - 1 [/itex] - identity
    [tex] (2 \cos^2 x - 1) - 5 \cos x + 3 = 0 [/tex]
    [tex] 2 \cos^2 x - 5 \cos x + 2 = 0 [/tex]
    [tex] a = 2 \; \; b = -5 \; \; c = 2 [/tex]
    [tex] \cos x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(2)}}{2(2)} [/tex]
    [tex] \cos x = \frac{5 \pm \sqrt{9}}{4} [/tex]
    [tex] \cos x = \frac{5 \pm 3}{4} [/tex]
    [tex] \boxed{\cos x = \frac{1}{2}} [/tex]
    [tex] \cos x = 2 \; \; \text{(no solution)} [/tex]
    :biggrin:
     
    Last edited: Apr 19, 2005
  8. Apr 18, 2005 #7

    dextercioby

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    U have to disregard [itex] \cos x=2 [/itex],okay?

    Daniel.
     
  9. Apr 18, 2005 #8

    Zurtex

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    Unless the question asks for complex solutions.
     
  10. Apr 18, 2005 #9

    dextercioby

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    "x" typically depicts real variable...:wink:

    If it were "z",it may have considered imaginary solutions,too...

    Daniel.
     
  11. Apr 19, 2005 #10
    Practice Problems...



    It is my opinion that if the original equation states both [itex] \sin x [/itex] and [itex] \cos x [/itex] then both solutions must be located in order to lock on the solution quadrants, if possible.

    [tex] 4 \sin^2 x + 2 \cos^2 x = 3 [/tex]
    [tex] \sin^2 x = 1 - \cos^2 x [/tex] - identity
    [tex] 4 (1 - \cos^2 x) + 2 \cos^2 x - 3 = 0 [/tex]
    [tex] 4 - 4 \cos^2 x + 2 \cos^2 x - 3 = 0 [/tex]
    [tex] -2 \cos^2 x + 1 = 0 [/tex]
    [tex] \cos^2 x = \frac{1}{2} [/tex]
    [tex] \boxed{\cos x = \pm \frac{\sqrt{2}}{2}} [/tex]
    [tex] \boxed{\cos x = \pm \frac{\sqrt{2}}{2} \; \; \sin x = \pm \frac{\sqrt{2}}{2}} [/tex]
    [tex] \boxed{x = \pm \frac{\pi}{4} \; \; x = \pm \frac{3 \pi}{4}} [/tex] - solution 1
    [tex] \boxed{x = \frac{\pi}{4} \; \; x = \frac{3 \pi}{4} \; \; x = \frac{5 \pi}{4} \; \; x = \frac{7 \pi}{4}} [/tex] - solution 2
    :uhh:

    By locking on the quadrants, a solution for [itex] x [/itex] can be found for 'special angles', without using inverse functions. However, in this equation, solutions exist in all four quadrants.

    A physics professor will accept solution 1 or 2 for [itex] x [/itex], however a mathematics professor will only accept solution 2 for [itex] x [/itex].
     
  12. Apr 19, 2005 #11

    dextercioby

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    Nope,as u can see,the initial equations placed no restrictions on the domains of the involved functions.So i'm sorry to dissapoint you,but u have search for solutions in [itex] \mbox{R} [/itex].

    Daniel.
     
  13. Apr 21, 2005 #12
    orion1
    u good ,dont get 2 impressed
     
  14. Apr 21, 2005 #13
    orion i still insist u good ,for the second trig question x=60 or 300 which u got right .see ya
     
  15. Apr 26, 2005 #14
    Range Rover...


    If the range for function 1 is [itex] \left[ 0 . 2 \pi \right] [/itex] then solution 1 and 2 are correct, however if the range for function 1 is [itex](-\infty . +\infty)[/itex] then the number of solutions for [itex]x[/itex] are infinite and the solution for [itex]R[/itex] becomes:

    [tex]x = \frac{\pi}{4} + \frac{n \pi}{2} = \frac{\pi}{2} \left( n + \frac{1}{2} \right)[/tex]
    [tex]\boxed{ x = \pm \frac{\pi}{2} \left( n + \frac{1}{2} \right)}[/tex]

    For function 2, the number of solutions for [itex]x[/itex] are infinite and limited solutions for [itex]x[/itex] becomes:
    [tex]x = \cos^{-1} \frac{1}{2}[/tex]
    [tex]\boxed{x = \pm \frac{\pi}{3}}[/tex]
    [tex]\boxed{x = \frac{\pi}{3} \; \; x = \frac{5 \pi}{3}}[/tex]

    For function 2, what is the [itex]x[/itex] solution for [itex]R[/itex]?
     
    Last edited: Apr 26, 2005
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