How do you solve #sqrt3cscx-2=0#?

1 Answer
Sep 5, 2015

#{x | x = pi/3 + 2kpi or x = (2pi)/3 + 2kpi, k in ZZ}#

Explanation:

To solve this equation, we need to isolate #x#, which means isolating the trigonometric function that has #x# in it.
If you find it hard to remember the unit circle values for secant or cosecant, you can always use the more familiar sine and cosine functions:

#sqrt(3)csc(x) - 2 = 0#

#sqrt(3)*1/sin(x) - 2 = 0#

#sqrt(3)*1/sin(x) = 2#

#sqrt(3) = 2sin(x)#

#sqrt(3)/2 = sin(x)#

The values for #x# where #sqrt(3)/2 = sin(x)# are #pi/3# and #(2pi)/3# radians.

Since there's no specified domain in the problem, the answer is the general solution.
The #sin# function has the period #2pi#, #pi/3# and #(2pi)/3# repeat every #2pi# in the answer.

#{x | x = pi/3 + 2kpi or x = (2pi)/3 + 2kpi, k in ZZ}#