Question

What are the solutions to the equation Sine (x + StartFraction 7 pi Over 2 EndFraction) = negative StartFraction StartRoot 3 EndRoot Over 2 EndFraction over the interval [0, 2Pi]?

Answer 1

Answer:

C. π/6 & 11π/6

Step-by-step explanation:

If you graph the equation ( Sin (x+7π/2)=-√3/2) and look between 0 & 2π, you'll see that the lines intersect the x-axis at π/6 & 11π/6.

Answer 2

Given:

The equation is

$\sin\left(x+\dfrac{7\pi}{2}\right)=-\dfrac{\sqrt{3}}{2}$

To find:

The solutions of given equation over the interval $[0,2\pi]$.

Solution:

We have,

The equation is

$\sin\left(x+\dfrac{7\pi}{2}\right)=-\dfrac{\sqrt{3}}{2}$

$\sin\left(x+\dfrac{7\pi}{2}\right)=-\sin \dfrac{\pi }{3}$

$\sin\left(x+\dfrac{7\pi}{2}\right)=\sin (-\dfrac{\pi }{3})$

If $\sin x=\sin y$, then $x=n\pi +(-1)^ny$.

Over the interval $[0,2\pi]$.

$x+\dfrac{7\pi}{2}=4\pi-\dfrac{\pi }{3}$ and $x+\dfrac{7\pi}{2}=5\pi+\dfrac{\pi }{3}$

$x=\dfrac{11\pi }{3}-\dfrac{7\pi}{2}$ and $x=\dfrac{16\pi}{3}-\dfrac{7\pi}{2}$

$x=\dfrac{22\pi-21\pi }{6}$ and $x=\dfrac{32\pi-21\pi }{6}$

$x=\dfrac{\pi}{6}$ and $x=\dfrac{11\pi }{6}$

Therefore, the two solutions are tex]x=\dfrac{\pi}{6}[/tex] and $x=\dfrac{11\pi }{6}$.