solve the equation for x, where x is restricted to the given interval. y = √7 + 2tan3x, for x in (-π/6, π/6)…

Explanation:

Step1: Isolate the tangent - term

Subtract $\sqrt{7}$ from both sides of the equation $y = \sqrt{7}+2\tan3x$. $y-\sqrt{7}=2\tan3x$

Step2: Solve for $\tan3x$

Divide both sides of the equation $y - \sqrt{7}=2\tan3x$ by 2. $\tan3x=\frac{y - \sqrt{7}}{2}$

Step3: Use the inverse - tangent function

Since $y = \tan x$ has an inverse function $y=\tan^{- 1}x$ for $x\in(-\frac{\pi}{2},\frac{\pi}{2})$, and $3x\in(-\frac{\pi}{2},\frac{\pi}{2})$ when $x\in(-\frac{\pi}{6},\frac{\pi}{6})$. $3x=\tan^{-1}(\frac{y - \sqrt{7}}{2})$

Step4: Solve for $x$

Divide both sides of the equation $3x=\tan^{-1}(\frac{y - \sqrt{7}}{2})$ by 3. $x=\frac{1}{3}\tan^{-1}(\frac{y - \sqrt{7}}{2})$

Answer:

B. $x = \frac{1}{3}\tan^{-1}(\frac{y-\sqrt{7}}{2})$