what are the solutions to the equation sin(x + \\frac{7\\pi}{2}) = -\\frac{\\sqrt{3}}{2} over the interval…

what are the solutions to the equation sin(x + \\frac{7\\pi}{2}) = -\\frac{\\sqrt{3}}{2} over the interval 0, 2\\pi?\n\\frac{\\pi}{6} only\n\\frac{\\pi}{6} and \\frac{5\\pi}{6} only\n\\frac{\\pi}{6} and \\frac{11\\pi}{6} only\n\\frac{\\pi}{6}, \\frac{5\\pi}{6}, \\frac{7\\pi}{6} and \\frac{11\\pi}{6}

what are the solutions to the equation sin(x + \\frac{7\\pi}{2}) = -\\frac{\\sqrt{3}}{2} over the interval 0, 2\\pi?\n\\frac{\\pi}{6} only\n\\frac{\\pi}{6} and \\frac{5\\pi}{6} only\n\\frac{\\pi}{6} and \\frac{11\\pi}{6} only\n\\frac{\\pi}{6}, \\frac{5\\pi}{6}, \\frac{7\\pi}{6} and \\frac{11\\pi}{6}

Answer

Explanation:

Step1: Use trigonometric identity

We know that $\sin(A + B)=\sin A\cos B+\cos A\sin B$. Also, $\sin\left(x+\frac{7\pi}{2}\right)=\sin x\cos\frac{7\pi}{2}+\cos x\sin\frac{7\pi}{2}$. Since $\cos\frac{7\pi}{2} = 0$ and $\sin\frac{7\pi}{2}=- 1$, then $\sin\left(x+\frac{7\pi}{2}\right)=-\cos x$. So the equation $\sin\left(x + \frac{7\pi}{2}\right)=-\frac{\sqrt{3}}{2}$ becomes $-\cos x=-\frac{\sqrt{3}}{2}$, or $\cos x=\frac{\sqrt{3}}{2}$.

Step2: Find solutions in the given interval

The general solution of the equation $\cos x = \frac{\sqrt{3}}{2}$ is $x = 2k\pi\pm\frac{\pi}{6},k\in\mathbb{Z}$. For the interval $[0,2\pi]$, when $k = 0$, $x=\frac{\pi}{6}$; when $k = 1$, $x = 2\pi-\frac{\pi}{6}=\frac{11\pi}{6}$.

Answer:

C. $\frac{\pi}{6}$ and $\frac{11\pi}{6}$ only