Asked by Anonymous
                csc^2 x = 5 over the interval 0 <= x < 2pi
1. 26.57 degs
2. 153.43 degs
I think the solutions in the third and fourth quadrants don't work because they don't fall in the interval.
            
        1. 26.57 degs
2. 153.43 degs
I think the solutions in the third and fourth quadrants don't work because they don't fall in the interval.
Answers
                    Answered by
            Reiny
            
    csc^2 x = 5
or
sin^2 x = 1/5
sin x = ± 1/√5
So x is in all 4 quadrants.
the reference angle is 26.57°
so x = 26.57°
x = 180-26.57 = 153.43°
x = 180 + 26.57 = 206.57°
or
x = 360-26.57 = 333.43°
just notice that your domain is 0 ≤ x ≤ 2π
so you would want your answers in radians,
x = .4636
x= π - .4636 = 2.68
x = π + .4636 = 3.61
x = 2π-.4636 = 5.82
    
or
sin^2 x = 1/5
sin x = ± 1/√5
So x is in all 4 quadrants.
the reference angle is 26.57°
so x = 26.57°
x = 180-26.57 = 153.43°
x = 180 + 26.57 = 206.57°
or
x = 360-26.57 = 333.43°
just notice that your domain is 0 ≤ x ≤ 2π
so you would want your answers in radians,
x = .4636
x= π - .4636 = 2.68
x = π + .4636 = 3.61
x = 2π-.4636 = 5.82
                    Answered by
            Anonymous
            
    Thank you so much!
    
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