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Math Help - solving trigonometric equation

  1. #1
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    solving trigonometric equation

    solve tanx+secx=2cosx
    (sinx/cosx)+ (1/cosx)=2cosx
    (sinx+1)/cosx =2cosx
    multiplying both sides by cosx
    sinx + 1 =2cos^2x
    sinx+1 = 2(1-sin^2x)
    2sin^2x + sinx-1=0
    (2sinx+1)(sinx-1)=0
    x=30 x=270
    but if i plug 270 back into the original equation i get undefined because tan 270 is undefined where did i go wrong?
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  2. #2
    Super Member Matt Westwood's Avatar
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    Because neither of tan x nor sec x are defined at 270 degrees, the original derivation of your quadratic in sines does not apply (good job btw) so it's not a valid solution.

    (Note you also have 30 + integer multiples of 360 degrees as solutions as well.)

    Try plotting tanx + secx and 2 cosx on the same graph and see what it looks like. Note the singular point at \pm 90 + 360n to see what I mean.
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  3. #3
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    so i do i solve it algebraically.
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  4. #4
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    Hello, myoplex11!

    Solve: . \tan x + \sec x\:=\:2\cos x

    \frac{\sin x}{\cos x} + \frac{1}{\cos x}\:=\:2\cos x

    Note that, at this point: \cos x \neq 0

    . . We cannot have zero in the denominator, right?

    That is: . x \:\neq\:\pm90^o,\;\pm270^o,\;\hdots


    So, should any of those values turn up later, they must be discarded.

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  5. #5
    Super Member Matt Westwood's Avatar
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    Beg pardon, was that a question?

    Yes, I would solve it algebraically like you did (it works, it's fine, as long as cos theta not zero) and then I'd probably plot some points and draw a sketch to make sure I didn't miss any points where tanx+secx intersect 2cosx.
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