<s>[INST] Easy way of tackling $\sqrt{x^2}+x=\sqrt{5}$ Solve for x
$$\sqrt{x^2}+x=\sqrt{5}\tag1$$
$$\sin(\sqrt{x^2}+x)=\sin({\sqrt{5}})\tag2$$
$$\sin(\sqrt{x^2})\cos(x)+\cos(\sqrt{x^2})\sin(x)=\sin(\sqrt{5})\tag3$$
$$\sin(\sqrt{x^2})+\cos(\sqrt{x^2})\tan(x)=\sin(\sqrt{5})\sec(x)\tag4$$
$$\sin(\sqrt{x^2})+\cos(\sqrt{x^2})\sqrt{\sec^2(x)-1}=\sin(\sqrt{5})\sec(x)\tag5$$
finally after all the hard work $(5)$ reveals
$$x+x=\sqrt{5}\tag6$$
$$x=\frac{\sqrt{5}}{2}\tag7$$
Is there another easy way solving this $(1)?$ 
 [/INST] Use cases. Certainly $x \neq 0.$
Case I: $x > 0.$ Then $\sqrt{x^2} + x = 2x.$ Thus $x = \frac{\sqrt{5}}{2}.$
Case II: $x < 0.$. Then $\sqrt{x^2} = -x$, but then this says $ 0 = \sqrt{5},$ contradiction. Hence we must have the only solution $x = \frac{\sqrt{5}}{2}.$
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