solve the equation for x, where x is restricted to the given interval. y = tan(x + 2), for x in (-π/2 - 2…

Explanation:

Step1: Apply inverse - tangent

Given $y = \tan(x + 2)$, we take the inverse - tangent of both sides. Since $x\in\left(-\frac{\pi}{2}-2,\frac{\pi}{2}-2\right)$, the inverse - tangent function $\arctan$ is well - defined. $\arctan(y)=\arctan(\tan(x + 2))$

Step2: Simplify the right - hand side

By the property of the inverse tangent function $\arctan(\tan(u))=u$ for $u\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, and here $u=x + 2$ with $x+2\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ (because $x\in\left(-\frac{\pi}{2}-2,\frac{\pi}{2}-2\right)$), we have: $\arctan(y)=x + 2$

Step3: Solve for x

Subtract 2 from both sides of the equation. $x=\arctan(y)-2$

Answer:

$x=\arctan(y)-2$