Question

find the values of x on the curve y = \frac{\cos(x)}{2 + \sin(x)} at which the tangent is horizontal. (use n as your integer variable. enter your answers as a comma - separated list.)

Explanation:

Step1: Recall the quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = \cos(x)$ and $v=2+\sin(x)$. So, $u^\prime=-\sin(x)$ and $v^\prime=\cos(x)$. Then $y^\prime=\frac{-\sin(x)(2 + \sin(x))-\cos(x)\cdot\cos(x)}{(2 + \sin(x))^{2}}$.

Step2: Simplify the derivative

Expand the numerator: $y^\prime=\frac{-2\sin(x)-\sin^{2}(x)-\cos^{2}(x)}{(2 + \sin(x))^{2}}$. Since $\sin^{2}(x)+\cos^{2}(x) = 1$, the numerator becomes $y^\prime=\frac{-2\sin(x)-1}{(2 + \sin(x))^{2}}$.

Step3: Set the derivative equal to zero

A horizontal tangent occurs when $y^\prime = 0$. Since the denominator $(2+\sin(x))^{2}\gt0$ for all real $x$ (because $- 1\leqslant\sin(x)\leqslant1$, so $1\leqslant2+\sin(x)\leqslant3$), we set the numerator equal to zero. So, $-2\sin(x)-1 = 0$.

Step4: Solve for $\sin(x)$

Rearrange the equation $-2\sin(x)-1 = 0$ to get $\sin(x)=-\frac{1}{2}$.

Step5: Find the values of $x$

The solutions of the equation $\sin(x)=-\frac{1}{2}$ are $x=\frac{7\pi}{6}+2n\pi$ and $x=\frac{11\pi}{6}+2n\pi$, where $n\in\mathbb{Z}$.

Answer:

$x=\frac{7\pi}{6}+2n\pi,\frac{11\pi}{6}+2n\pi$