solve: $cos(x+pi)=\frac{1}{2}$ over the interval $\frac{pi}{2},pi$.\n$\frac{pi}{2}$\n$\frac{2pi}{3}$\n$\frac{…

solve: $cos(x+pi)=\frac{1}{2}$ over the interval $\frac{pi}{2},pi$.\n$\frac{pi}{2}$\n$\frac{2pi}{3}$\n$\frac{3pi}{4}$\n$\frac{5pi}{6}$

solve: $cos(x+pi)=\frac{1}{2}$ over the interval $\frac{pi}{2},pi$.\n$\frac{pi}{2}$\n$\frac{2pi}{3}$\n$\frac{3pi}{4}$\n$\frac{5pi}{6}$

Answer

Explanation:

Step1: Use cosine addition formula

We know that $\cos(A + B)=\cos A\cos B-\sin A\sin B$. So, $\cos(x+\pi)=\cos x\cos\pi-\sin x\sin\pi$. Since $\cos\pi=- 1$ and $\sin\pi = 0$, then $\cos(x+\pi)=-\cos x$. The equation $\cos(x + \pi)=\frac{1}{2}$ becomes $-\cos x=\frac{1}{2}$, or $\cos x=-\frac{1}{2}$.

Step2: Find x in given interval

We know that $\cos x =-\frac{1}{2}$ has solutions $x = 2k\pi\pm\frac{2\pi}{3},k\in\mathbb{Z}$. In the interval $[\frac{\pi}{2},\pi]$, when we consider the general - solution and substitute $k = 0$, the solution of $\cos x=-\frac{1}{2}$ in this interval is $x=\frac{2\pi}{3}$.

Answer:

B. $\frac{2\pi}{3}$