Question

Find `arcsin(-sqrt3/2)`?

Answer 1

`arcsin(-sqrt3/2)` is equal to `(2n+1)pi+-pi/3`, where `n` is an integer

Explanation:

`arcsin(-sqrt3/2)` is the angle whose sine is `-sqrt3/2`.

As sine is negative in second and third quadrant and `sin(pi/3)=sqrt3/2`,

we have `sin(pi-pi/3)=-sqrt3/2` and

`sin(p+pi/3)=-sqrt3/2`

Hence `arcsin(-sqrt3/2)` is equal to `(2pi)/3` or `(4pi)/3`.

As adding or subtracting `2pi` does not affect trigonometric ratios of angles, we can have infinite solutions given by

`(2n+1)pi+-pi/3`, where `n` is an integer.

Answer 2

`npi+(-1)^n(-pi/3)`, n = 0, 1, 2, 3,...

Explanation:

If sin x = a and `alpha` is the principal value of `x in [-pi/2, pi/2]`, then the general value of x = `npi+(-1)^nalpha`, n = 0, 1, 2, 3,...

Here `x = arc sin (-sqrt3/2)`
`sin x = -sqrt3/2`.
So, `alpha = -pi/3`..

The answer is as stated.

Answer 3

`x = (4pi)/3 + 2kpi`
`x = (5pi)/3 + 2kpi`

Explanation:

`sin x = (-sqrt3/2)`. Find arc x (or angle x)
Unit circle gives 2 solutions for arc x (or angle x) -->
`x = - pi/3 + 2kpi`, and
`x = pi - (-pi/3) = pi + pi/3 = (4pi)/3 + 2kpi`
Note that `x = -pi/3` is co-terminal to `x = (5pi)/3.`
Answers:
`x = (4pi)/3 + 2kpi`
`x = (5pi)/3 + 2kpi`