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

what are the solutions to the equation ( sin left( x + \frac { 7 pi } { 2 } \right) = - \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 left( x + \frac { 7 pi } { 2 } \right) = - \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 the trigonometric identity

We know that (\sin(A)=-\frac{\sqrt{3}}{2}) when (A = \frac{4\pi}{3}+ 2k\pi) or (A=\frac{5\pi}{3}+2k\pi,k\in\mathbb{Z}). Here (A=x + \frac{7\pi}{2}). So (x+\frac{7\pi}{2}=\frac{4\pi}{3}+2k\pi) or (x+\frac{7\pi}{2}=\frac{5\pi}{3}+2k\pi).

Step2: Solve for (x) in the first - case

For (x+\frac{7\pi}{2}=\frac{4\pi}{3}+2k\pi): [ \begin{align*} x&=\frac{4\pi}{3}+2k\pi-\frac{7\pi}{2}\ &=\frac{8\pi - 21\pi}{6}+2k\pi\ &=-\frac{13\pi}{6}+2k\pi \end{align*} ] When (k = 1), (x=-\frac{13\pi}{6}+2\pi=\frac{-13\pi + 12\pi}{6}=\frac{\pi}{6})

Step3: Solve for (x) in the second - case

For (x+\frac{7\pi}{2}=\frac{5\pi}{3}+2k\pi): [ \begin{align*} x&=\frac{5\pi}{3}+2k\pi-\frac{7\pi}{2}\ &=\frac{10\pi-21\pi}{6}+2k\pi\ &=-\frac{11\pi}{6}+2k\pi \end{align*} ] When (k = 1), (x=-\frac{11\pi}{6}+2\pi=\frac{-11\pi + 12\pi}{6}=\frac{11\pi}{6})

Answer:

(\frac{\pi}{6}) and (\frac{11\pi}{6}) only