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 $(1)?$
There is another way to solve this.

Using the fact that $\sqrt{x^2} = |x|$. Your equation turns out to be $$|x|+x=\sqrt{5}$$

First look for solutions where $x\geq 0$ in this case you get $x+x=\sqrt{5}$ and so $x=\frac{\sqrt{5}}{2}$.

Now if $x&lt;0$ then $|x|=-x$ and so you get $-x+x=\sqrt{5}$ and clearly no $x$ satisfies this condition.
Use cases. Certainly $x \neq 0.$

Case I: $x &gt; 0.$ Then $\sqrt{x^2} + x = 2x.$ Thus $x = \frac{\sqrt{5}}{2}.$

Case II: $x &lt; 0.$. Then $\sqrt{x^2} = -x$, but then this says $ 0 = \sqrt{5},$ contradiction. Hence we must have the only solution $x = \frac{\sqrt{5}}{2}.$