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Solve the trigonometric equation for the values of x between 0 and 360

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csc²x-4=0

I don't really know how to approach this question. I'm working ahead in my textbook and decided to give it a shot. All help is appreaciated :)

 Feb 9, 2019

4+0 Answers

 #1
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csc = 1/sin

csc^2 = 1/sin^2

 

1/ sin^2 = 4

sin^2 = 1/4

sin = +- 1/2        Got it from here?

 Feb 9, 2019
edited by ElectricPavlov  Feb 9, 2019
 #2
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So with sin^2x = 1/4 how do i find the values of x between 0 and 360 degrees

Guest Feb 9, 2019
 #3
avatar+36916 
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arcsin (1/2)  =   30     150

 

 

arcsin (-1/2) =   330        210 degrees     

 

Sorry! (for the edits)

ElectricPavlov  Feb 9, 2019
edited by ElectricPavlov  Feb 9, 2019
edited by ElectricPavlov  Feb 9, 2019
 #4
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It's ok about the edits! I figured out what you were trying to say. And then I just use the 4 quadrants im assuming (theta+180, -180 and -360)

Guest Feb 9, 2019

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