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

C. π/6 & 11π/6

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+(7\pi)/(2)\right)=-(√(3))/(2)`

To find:

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

Solution:

We have,

The equation is

`\sin\left(x+(7\pi)/(2)\right)=-(√(3))/(2)`

`\sin\left(x+(7\pi)/(2)\right)=-\sin (\pi )/(3)`

`\sin\left(x+(7\pi)/(2)\right)=\sin (-(\pi )/(3))`

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

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

`x+(7\pi)/(2)=4\pi-(\pi )/(3)` and
`x+(7\pi)/(2)=5\pi+(\pi )/(3)`

`x=(11\pi )/(3)-(7\pi)/(2)` and
`x=(16\pi)/(3)-(7\pi)/(2)`

`x=(22\pi-21\pi )/(6)` and
`x=(32\pi-21\pi )/(6)`

`x=(\pi)/(6)` and
`x=(11\pi )/(6)`

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