how do you solve cscx+ 2=0 on the interval (0,2pi)

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jeew-m | College Teacher | (Level 1) Educator Emeritus

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`cscx +2 = 0`

`cscx = -2`

`1/sinx = -2`

`sinx = -1/2`

 

`sinx = -1/2`

`sinx = -sin(pi/6)`

`sinx = sin(-pi/6)`

 

The common solution for sines is` x = npi+(-1)^nalpha`

 

`sinx = sin(-pi/6)`

    `x = npi+(-1)^n(-pi/6) ` where `n in Z`

 

n                    x

0                 ` -pi/6`  

1                  `7pi/6 `

2                  `11pi/6 `

3                 `19pi/6`

 

So the answers in the interval `(0,2pi) ` will be;

`x = (7pi)/6 `

`x = (11pi)/6 `

 

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