Solving a trigonometric equation with multiples of ##\tan##

[brotherbobby]

Homework Statement

Find the smallest positive value of $x$ in degrees for which $\boxed{\tan(x+100^{\circ})=\tan(x+50^{\circ})\tan x\tan(x-50^{\circ}){\color{red}{\quad (x=?)}}}$

Relevant Equations

1. Trigonometry : $\small{\textbf{(a)}\;2\sin A \cos B = \sin(A+B)+\sin(A-B) \;\;\textbf{(b)}\; \sin C-\sin D = 2\cos\dfrac{C+D}{2}\sin\dfrac{C-D}{2}}$

2. Algebra : The componendo and dividendo $\quad\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow \dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}$

Attempt :

$\begin{align}\tan(x+100^{\circ})&=\tan(x+50^{\circ})\tan x\tan(x-50^{\circ})\\ \Rightarrow \dfrac{\sin(x+100^{\circ})\cos x}{\cos(x+100^{\circ})\sin x}&=\dfrac{\sin(x+50^{\circ})\cos (x-50^{\circ})}{\cos(x+50^{\circ})\sin (x-50^{\circ})}\quad{\text{(Using tan x=sin x/cos x)}}\\ \Rightarrow\dfrac{\sin(2x+100^{\circ})+\sin 100^{\circ}}{\sin(2x+100^{\circ})-\sin 100^{\circ}} &= \dfrac{\sin2x+\sin 100^{\circ}}{\sin2x-\sin 100^{\circ}} \qquad\qquad\quad{\text{(Using 1 (a) above)}} \\ \Rightarrow \dfrac{\cancel{2} \sin(2x+100^{\circ})}{\cancel{2}\sin 100^{\circ}}&= \dfrac{\cancel{2}\sin 2x}{\cancel{2}\sin 100^{\circ}}\qquad\qquad\qquad\qquad{\text{(Using 2 above)}}\\ \Rightarrow \sin(2x+100^{\circ})-\sin 2x&=0\\ 2\cos(2x+50^{\circ})\sin 50^{\circ} &= 0\qquad\qquad\qquad\qquad\qquad\qquad{\text{(Using 1(b) above)}}\\ \Rightarrow 2x+50^{\circ}&=90^{\circ}\\ \Rightarrow \mathbf {x} &= \mathbf{20^{\circ}} \quad {\color{red}{\Huge{\times}}} \end{align}$

Answer from text : The text gives the answer as $\mathbf{x=30^{\circ}}\quad{\color{DarkGreen}{\Huge{\checkmark}}}$

Request : The text answer matches the L.H.S. and the R.H.S. of the given equation. I'd like to know where have I gone wrong.

 

[Ibix]

Right hand side of your second line. You seem to have substituted cos/sin for the $\tan(x-50)$ term instead of sin/cos.

 

[brotherbobby]

Thanks @Ibix and sorry for the late response as well as the silly error. The trouble is, despite correcting myself, I am no closer to the answer as to what $x=?$

I do the calculation below using $\text{Autodesk Sketchbook}^{\circledR}$, hoping am not violating anything.

Problem statement : Find the smallest positive value of $x$ in degrees for which $\small{\boxed{\tan(x+100^{\circ})=\tan(x+50^{\circ})\tan x\tan(x-50^{\circ}){\color{red}{\quad (x=?)}}}}$

Attempt :

I'd like help as to how to proceed from here. Perhaps I have gone wrong at the first step?

 

[Ibix]

I must say I can't see it either. I'll have a think.

 

[Gavran]

I'd like help as to how to proceed from here.

$\begin{align} \sin(4x+100^\circ)&=-\sin100^\circ-\sin200^\circ\nonumber\\ &=\sin260^\circ-\sin200^\circ\nonumber\\ \end{align}$

 

[brotherbobby]

$\begin{align} \sin(4x+100^\circ)&=-\sin100^\circ-\sin200^\circ\nonumber\\ &=\sin260^\circ-\sin200^\circ\nonumber\\ \end{align}$

Sorry for coming in late. I haven't been able to solve the problem despite your help.

The difference of two sines : $\sin C - \sin D = 2 \cos\dfrac{C+D}{2}\sin\dfrac{C-D}{2}$.

Thus, from the last line of your post above, we have $\sin A = 2 \cos\dfrac{C+D}{2}\sin\dfrac{C-D}{2}$.

How do I proceed from here?

 

[Gavran]

Sorry for coming in late. I haven't been able to solve the problem despite your help.

The difference of two sines : $\sin C - \sin D = 2 \cos\dfrac{C+D}{2}\sin\dfrac{C-D}{2}$.

Thus, from the last line of your post above, we have $\sin A = 2 \cos\dfrac{C+D}{2}\sin\dfrac{C-D}{2}$.

How do I proceed from here?

Calculate $$ \sin\frac{C-D}{2} $$ and use $$ \cos\frac{C+D}{2}=\sin(450^\circ-\frac{C+D}{2}) $$.

 

[renormalize]

How do I proceed from here?

I haven't verified your derivation, but assuming your equation:

is correct, I would proceed as follows: define $y\equiv 4x, \theta\equiv 100°$ so that your equation becomes:
\begin{align}
0 & =\sin\left(y+\theta\right)+\sin\theta+\sin2\theta \nonumber \\
& = \cos y\sin\theta+\sin y\cos\theta+\sin\theta+\sin2\theta \nonumber
\end{align}
where in the second line I've expanded $\sin\left(y+\theta\right)$. Can you take it from there to solve for $y$ (and hence $x$) as a function of $\theta\,$?