Find the value of $x$, if both angles are acute and $\cot (3x-40) = \tan (2x+100)$

Question

Find the value of $x$, if both angles are acute and $\cot (3x-40) = \tan (2x+100)$

All I know is that $\tan (x+90) = - \cot(x)$. However, I didn't know how to apply that into this question.

Answer 1

$$\cot(3x - 40) = \tan(90 - 3x + 40) = \tan(130-3x)$$ [As the angle $3x - 40$ is acute, ($<90$)]

$$ \Rightarrow \tan(2x+100)=\tan(130-3x) $$

$$ \Rightarrow 2x+100 = n(180) +130 - 3x $$

$$\Rightarrow 5x = n(180)+30$$

$$ \boxed {x = 6 +n(180) \deg = \frac{\pi}{30}+n\pi \ [n \in \Bbb Z]}$$