which of the following are solutions to the equation sinx cosx = -1/4? check all that apply. a. 11π/12 + nπ…

Answer:

A. $\frac{11\pi}{12}+n\pi$, B. $\frac{7\pi}{12}+n\pi$

Explanation:

Step1: Use double - angle formula

Recall $\sin x\cos x=\frac{1}{2}\sin(2x)$. So the equation $\sin x\cos x =-\frac{1}{4}$ becomes $\frac{1}{2}\sin(2x)=-\frac{1}{4}$.

Step2: Solve for $\sin(2x)$

Multiply both sides of $\frac{1}{2}\sin(2x)=-\frac{1}{4}$ by 2 to get $\sin(2x)=-\frac{1}{2}$.

Step3: Find the general solutions for $2x$

The general solutions of $\sin\theta =-\frac{1}{2}$ are $\theta=\frac{7\pi}{6}+2k\pi$ or $\theta=\frac{11\pi}{6}+2k\pi$, $k\in\mathbb{Z}$. So $2x=\frac{7\pi}{6}+2k\pi$ or $2x=\frac{11\pi}{6}+2k\pi$.

Step4: Solve for $x$

For $2x=\frac{7\pi}{6}+2k\pi$, divide by 2 to get $x = \frac{7\pi}{12}+k\pi$. For $2x=\frac{11\pi}{6}+2k\pi$, divide by 2 to get $x=\frac{11\pi}{12}+k\pi$. Here $n = k$.