Question

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

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 \(x+\frac{7\pi}{2}=\frac{4\pi}{3}+2k\pi\):

$$
LATEXBLOCK0
$$

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 \(x+\frac{7\pi}{2}=\frac{5\pi}{3}+2k\pi\):

$$
LATEXBLOCK1
$$

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