# Solve for $x$ in $-\arctan(0.3x)-2x=-180^{\circ}$?

How can I solve for $x$ in the following: $$-\arctan(0.3x)-2x=-180^{\circ} \qquad ?$$ I tried \begin{align} \arctan(0.3x)&=180^{\circ}-2x \\ 0.3x&=\tan(180^{\circ}-2x)\\ \end{align} With the identity $\tan(\alpha\pm\beta)=\frac{\tan\alpha\pm\tan\beta}{1\mp\tan\alpha \tan \beta }$, I have \begin{align} 0.3x&=\tan(180^{\circ}-2x)\\ &=\frac{\tan 180^{\circ}- \tan 2x}{1+\tan 180^{\circ} \, \tan 2x}\\ &=-\tan 2x \end{align} So I'm stuck with $0.3x=-\tan 2x$, is it correct? How should I proceed?

• You cannot go further analytically. – Szeto Jul 17 '18 at 12:53
• a trivial solution is $x=0$ – Vasya Jul 17 '18 at 12:56
• $x=\tan x$ is transcendental, meaning the solution cannot be expressed in a finite sequence of algebraic operations. The best you can do is approximate – John Glenn Jul 17 '18 at 16:02

Let $$f(x)=\tan(2x) + 0.3x$$ and apply Newton's method to approximate zero's of $f(x).$