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Please show a method of solving the equation $\tan(x)=\cos(x+33.44)$.

I tried several methods (half-angle, cosine of sum, multiply cosines,etc...), but nothing worked.

How should one solve such equation or in general an equation of the form $\tan(x)=\sin(x+a)$ or $\tan(x)=\cos(x+a)$?

Thanks!

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Hint

$$\tan(\arccos(x))=\frac{x}{\sqrt{1-x^2}},$$

$$\sin(\arccos(x))=\sqrt{1-x^2}.$$

Set $x=\arccos(\alpha)$ and your equation, and use the formula above and the fact that $$\cos(a+b)=\cos(a)+\cos(b)-\sin(a)\sin(b)$$ to conclude.

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If all else fails you can use the substitution $t=\tan \frac x2$ for this kind of problem.

Then $$\tan x = \cos x\cos \alpha -\sin x \sin \alpha$$ becomes $$\frac {2t}{1-t^2}=\frac {1-t^2}{1+t^2}\cos \alpha-\frac {2t}{1+t^2}\sin \alpha$$ and clearing fractions gives $$2t+2t^3=(1-2t^2+t^4)\cos \alpha-(2t-2t^3)\sin \alpha$$ and this yields the quartic $$t^4\cos\alpha + 2t^3(\sin\alpha-1)-2t^2\cos \alpha-2t(\sin \alpha +1)+\cos \alpha=0$$

Which can be solved using any convenient method.

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  • $\begingroup$ I tried this method, but couldn't solve the quartic without numeric estimation or closed formula. Is there any easier way? $\endgroup$ – Galc127 Aug 12 '15 at 9:47
  • $\begingroup$ @Galc127 All the suggested methods seem to land with a quartic, and if you look at $\tan x=a\cos (x+b)$ for suitable $b$ and sufficiently large $a$ you do see four real solutions, so it looks like it can't be avoided. $\endgroup$ – Mark Bennet Aug 12 '15 at 10:14

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