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]?

Respuesta :

Given:

The equation is

[tex]\sin\left(x+\dfrac{7\pi}{2}\right)=-\dfrac{\sqrt{3}}{2}[/tex]

To find:

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

Solution:

We have,

The equation is

[tex]\sin\left(x+\dfrac{7\pi}{2}\right)=-\dfrac{\sqrt{3}}{2}[/tex]

[tex]\sin\left(x+\dfrac{7\pi}{2}\right)=-\sin \dfrac{\pi }{3}[/tex]

[tex]\sin\left(x+\dfrac{7\pi}{2}\right)=\sin (-\dfrac{\pi }{3})[/tex]

If [tex]\sin x=\sin y[/tex], then [tex]x=n\pi +(-1)^ny[/tex].

Over the interval [tex][0,2\pi][/tex].

[tex]x+\dfrac{7\pi}{2}=4\pi-\dfrac{\pi }{3}[/tex] and [tex]x+\dfrac{7\pi}{2}=5\pi+\dfrac{\pi }{3}[/tex]

[tex]x=\dfrac{11\pi }{3}-\dfrac{7\pi}{2}[/tex] and [tex]x=\dfrac{16\pi}{3}-\dfrac{7\pi}{2}[/tex]

[tex]x=\dfrac{22\pi-21\pi }{6}[/tex] and [tex]x=\dfrac{32\pi-21\pi }{6}[/tex]

[tex]x=\dfrac{\pi}{6}[/tex] and [tex]x=\dfrac{11\pi }{6}[/tex]

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

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.

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