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Thread: solve for x algebraically

  1. #1
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    solve for x algebraically

    solve x algebraically not graphically

    3 cot (90 + x) = tan x sin x

    Thanx
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  2. #2
    Super Member wingless's Avatar
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    $\displaystyle 3\cot (90+x) = \tan x \sin x$

    $\displaystyle 3\frac{\cos (90 + x)}{\sin (90 + x)} = \frac{\sin x}{\cos x} \sin x$

    $\displaystyle 3\frac{-\sin x}{\cos x} = \frac{\sin^2 x}{\cos x}$

    $\displaystyle 3\frac{-\sin x}{\not{\cos x}} = \frac{\sin^2 x}{\not{\cos x}}$

    $\displaystyle -3\sin x = \sin^2 x$

    .........$\displaystyle \sin x = 0$ works

    We'll try to get another answer,

    $\displaystyle -3\not{\sin x} = \sin^{\not 2} x$

    $\displaystyle -3 = \sin x$ There's no solution here.

    Then we only use $\displaystyle \sin x = 0$.

    Hence, $\displaystyle x = k\pi$
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  3. #3
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    Hello, Nico!

    A slightly different approach . . .


    $\displaystyle 3\cot(90 + x) \:=\: \tan x\sin x$

    Since $\displaystyle \cot(x + 90) \:=\:-\tan x$, we have:

    . . $\displaystyle -3\tan x \:=\:\tan x\sin x \quad\Rightarrow\quad \tan x\sin x + 3\tan x \:=\:0$

    Factor: .$\displaystyle \tan x(\sin x + 3) \:=\:0$


    But .$\displaystyle \sin x + 3 \:=\:0\quad\Rightarrow\quad \sin x \:=\:-3\;\text{ has no solutions}$

    Therefore: .$\displaystyle \tan x \:=\:0\quad\Rightarrow\quad x \:=\:\pi n\:\text{ for any integer }n$

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