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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?

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  • $\begingroup$ You cannot go further analytically. $\endgroup$ – Szeto Jul 17 '18 at 12:53
  • $\begingroup$ a trivial solution is $x=0$ $\endgroup$ – Vasya Jul 17 '18 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 '18 at 16:02
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Let $$f(x)=\tan(2x) + 0.3x$$ and apply Newton's method to approximate zero's of $f(x).$

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  • $\begingroup$ Make sure to graph the two functions first do you have a general idea of where the solutions are. $\endgroup$ – Shrey Joshi Jul 25 '18 at 13:33

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