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?

  • $\begingroup$ You cannot go further analytically. $\endgroup$
    – Szeto
    Jul 17, 2018 at 12:53
  • $\begingroup$ a trivial solution is $x=0$ $\endgroup$
    – Vasili
    Jul 17, 2018 at 12:56
  • $\begingroup$ $x=\tan x$ is transcendental, meaning the solution cannot be expressed in a finite sequence of algebraic operations. The best you can do is approximate $\endgroup$
    – John Glenn
    Jul 17, 2018 at 16:02

1 Answer 1


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

  • $\begingroup$ Make sure to graph the two functions first do you have a general idea of where the solutions are. $\endgroup$ Jul 25, 2018 at 13:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.