Solve for x in the interval [-pi,0]

Question

a) sin^2x = 3/4

I know that you have +root3/2 and
-root3/2 and the positive one gives you an error when doing the inverse of sin, but im confused about the -root3/2.

I found that one of the answers of x is
-pi/3 (-60 degrees) which fits my interval, but I don't know how you get
-2pi/3 as the other answer.

I did -pi/3 + pi but -2pi/3 as a positive angle which does not fit into the interval.

Please explain how you get the answer -2pi/3?

Answer 1

as you say, sinx = ±√3/2
You know the reference angle is π/3.

sinx > 0 in QI, QII, so you have π/3 and 2π/3
sinx < 0 in QIII, QIV, so that gives you 4π/3 and 5π/3

But, for this problem, the domain is [-π,0] (QIII and QIV), so you have to subtract 2π giving you -π/3 instead of 5π/3, and -2π/3 instead of 4π/3.

Answer 2

Trigonometric identity:

sin ( - x ) = - sin x

On interval [ 0 , pi ] [ 0 , 180 ° ]

you have 2 angles with sin x = sqroot ( 3 ) / 2

x = pi / 3 = 60 °

and

x = 2 pi / 3 = 120 °

So if sin ( - x ) = sin x

on interval [ - pi , 0 ] [ - 180 ° , 0 ]

you have 2 angles with sin x = sqroot ( 3 ) / 2

x = - pi / 3 = - 60 °

and

x = - 2 pi / 3 = - 120 °