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    Question determine all te possible value of x where 0 degree <= x <= 360 degree such that

    tan 2x = 6 cot x.

    -is it change the tan 2x or 6 cot x first?
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    Re: determine all te possible value of x where 0 degree <= x <= 360 degree such that

    Quote Originally Posted by choi105 View Post
    tan 2x = 6 cot x.

    -is it change the tan 2x or 6 cot x first?
    Ummm....Use the double angle formula to replace tan(2x). I think that's what you were asking.

    -Dan
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    Re: determine all te possible value of x where 0 degree <= x <= 360 degree such that

    \displaystyle \begin{align*} \tan{(2x)} &= 6\cot{(x)} \\ \frac{\sin{(2x)}}{\cos{(2x)}} &= \frac{6\cos{(x)}}{\sin{(x)}} \\ \frac{2\sin{(x)}\cos{(x)}}{\cos^2{(x)} - \sin^2{(x)}} &= \frac{6\cos{(x)}}{\sin{(x)}} \\ \frac{2\sin^2{(x)}\cos{(x)}}{\cos^2{(x)} - \sin^2{(x)}} &= 6\cos{(x)} \\ \frac{2\sin^2{(x)}\cos{(x)}}{\cos^2{(x)} - \sin^2{(x)} } - 6\cos{(x)} &= 0 \\ 2\cos{(x)} \left[ \frac{\sin^2{(x)}}{\cos^2{(x)} - \sin^2{(x)}} - 3 \right] &= 0 \\ \cos{(x)} = 0 \textrm{ or } \frac{\sin^2{(x)}}{\cos^2{(x)} - \sin^2{(x)}} - 3 &= 0  \end{align*}

    Case 1:

    \displaystyle \begin{align*} \cos{(x)} &= 0 \\ x &= \left( 2m + 1 \right) 90^{\circ} \textrm{ where } m \in \mathbf{Z} \end{align*}

    Case 2:

    \displaystyle \begin{align*} \frac{\sin^2{(x)}}{\cos^2{(x)} - \sin^2{(x)}} - 3 &= 0 \\ \frac{\sin^2{(x)}}{\cos^2{(x)} - \sin^2{(x)}} &= 3 \\ \sin^2{(x)} &= 3 \left[ \cos^2{(x)} - \sin^2{(x)} \right] \\ \sin^2{(x)} &= 3 \left[ 1 - 2\sin^2{(x)} \right] \\ \sin^2{(x)} &= 3 - 6\sin^2{(x)} \\ 7\sin^2{(x)} &= 3 \\ \sin^2{(x)} &= \frac{3}{7} \\ \sin{(x)} &= \pm \sqrt{ \frac{3}{7}} \\ \sin{(x)} &= \pm \frac{\sqrt{21}}{7} \\ x &= \left\{ \arcsin{ \left( \frac{\sqrt{21}}{7} \right) } ^{\circ} , 180^{\circ} - \arcsin{ \left( \frac{\sqrt{21}}{7} \right) } ^{\circ} , 180^{\circ} + \arcsin{ \left( \frac{\sqrt{21}}{7} \right) } ^{\circ} , 360^{\circ} - \arcsin{ \left( \frac{\sqrt{21}}{7} \right) } ^{\circ} \right\} + 360^{\circ} n \textrm{ where } n \in \mathbf{Z} \end{align*}

    So in the domain \displaystyle \begin{align*} 0^{\circ} \leq x \leq 360^{\circ} \end{align*}, we have

    \displaystyle \begin{align*} x = \left\{ \arcsin{ \left( \frac{\sqrt{21}}{7} \right) }^{\circ} , 90^{\circ}, 180^{\circ} - \arcsin{ \left( \frac{\sqrt{21}}{7} \right) } ^{\circ} , 180^{\circ} + \arcsin{ \left( \frac{\sqrt{21}}{7} \right) } ^{\circ} , 270^{\circ} , 360^{\circ} - \arcsin{ \left( \frac{\sqrt{21}}{7} \right) } ^{\circ} \right\} \end{align*}
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    Re: determine all te possible value of x where 0 degree <= x <= 360 degree such that

    Thank you very much.
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    Re: determine all te possible value of x where 0 degree <= x <= 360 degree such that

    Hello, choi105!

    \text{Solve for }x\text{ on }[0^o,\,360^o]\!:\;\tan 2x \:=\: 6\cot x

    We have: . \frac{2\tan x}{1-\tan^2\!x} \;=\;\frac{6}{\tan x} \quad\Rightarrow\quad 2\tan^2\!x \;=\;6 - 6\tan^2\!x

    . . 8\tan^2\!x \:=\:6 \quad\Rightarrow\quad \tan^2\!x \:=\:\frac{3}{4} \quad\Rightarrow\quad \tan x \:=\:\pm\frac{\sqrt{3}}{2}

    Hence: . x \;=\;\tan^{-1}\!\left(\tfrac{\sqrt{3}}{2}\right) \;=\;40.89339465^o \;\approx\;40.9^o

    Therefore: . x \;\approx\;\begin{Bmatrix}40.9^o \\ 139.1^o \\ 220.9^o \\ 319.1^o \end{Bmatrix}
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