Question

Choose the values that would complete the general solution for y = arcsin(-sqrt3/2). ___±2k π

Answer 1

The general solution of y = arcsin(-√3/2) includes angles -π/3 and -2π/3 adjusted by ± 2kπ for integer values of k to take into account the periodicity of the sine function.

To solve for the general solution of y = arcsin(-√3/2), one must find the angles whose sine is -√3/2.

The arcsine function (arcsin) is defined to only return values between -π/2 and π/2 (or -90° and 90°).

In this range, the sine of π/3 (or 60°) is √3/2. Nevertheless, since we are considering the negative value (-√3/2), the angle in the principal range would be -π/3 (or -60°).

Thus, the general solution for y involves the angles which are symmetric to -π/3 about the x-axis in the unit circle.

Considering the periodicity of the sine function which is 2π, the general solution for the equation can be written as:

y₁ = -π/3 ± 2kπ

y₂ = -2π/3 + π ± 2kπ (since the sine function is also negative at 7π/3 or -2π/3 in the unit circle)

Here k is any integer, representing the number of complete cycles the sine wave has undergone.

Answer 2

To solve this problem you must apply the proccedure shown below:
1. You have the following inverse trigonometric function given in the problem above:

`y=arcsin(- √(3) /2)`
2. By definition, you have that
`y=arcsinx` is equivalent to
`siny=x` and
`sin(-x)=-sin(x)`, therefore:

`-siny=( √(3)/2)`
4. The value of the sine of
`√(3)/2` is
`\pi/3`, therefore:

`y=- \pi/3`
The answer is:
`- \pi/3`