How do you solve #cscG= 2.3456#?

1 Answer
Alan P.
Oct 7, 2015

Within the range #[0,2pi)#
#G=0.440432 " (radians)"# or #pi-0.440432 " (radians)"#

Explanation:

#csc(G) = 2.3456#

#rarrcolor(white)("XXX")sin(G)=1/2.3456#

#arcsin(sin(G)) = G# if we restrict #G# to #(-pi/2,pi/2]#

#arcsin(1/2.3456) = 0.440432# (using a calculator)

Within the full #[0,2pi]# range an alternate possibility exists in Q II for which the #sin# has the same value; namely
#color(white)("XXX")pi - 0.440432#

Note: if you need to remove the #[0,2pi]# limitation allow #(+2npi, AAn in ZZ)# for both possible solutions

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