How do you solve #2csc^2 x + cot x = 10# in #[0^@, 360^@)# ?

Question

Dylan Arnold · Accepted Answer

#cot^(-1) ((1+sqrt(65))/4) ~~ 23.816^@#
#cot^(-1) ((1-sqrt(65))/4) ~~ 150.473^@#
#cot^(-1) ((1+sqrt(65))/4) + 180^@ ~~ 203.816^@#
#cot^(-1) ((1-sqrt(65))/4) + 180^@ ~~ 330.473^@#

Note that: #csc^2 x = 1/sin^2 x = (cos^2 x+sin^2 x)/sin^2 x = 1+cot^2 x# So: #10 = 2csc^2 x + cot x# #color(white)(10) = 2(cot^2 x + 1) + cot x# #color(white)(10) = 2cot^2 x + cot x + 2# Subtract #10# from both ends to get: #2cot^2 x + cot x - 8 = 0# This is in the form: #at^2+bt+c = 0# with #t = cot x#, #a=2#, #b=1# and #c=-8#. So using the quadratic formula, we have: #cot x = (-b+-sqrt(b^2-4ac))/(2a)# #color(white)(cot x) = (-color(blue)(1)+-sqrt(color(blue)(1)^2-4(color(blue)(2))(color(blue)(-8))))/(2(color(blue)(2)))# #color(white)(cot x) = (-1+-sqrt(65))/4# Note that (expressed in degrees): #cot x# is continuous and monotonically decreasing on #(0^@, 180^@)# with range #(-oo, oo)#. The range of #cot^(-1)(y)# is #(0^@, 180^@)#. #cot x# has period #180^@#. Hence all the solutions in #(0^@, 360^@)# are: #cot^(-1) ((1+sqrt(65))/4) ~~ 23.816^@# #cot^(-1) ((1-sqrt(65))/4) ~~ 150.473^@# #cot^(-1) ((1+sqrt(65))/4) + 180^@ ~~ 203.816^@# #cot^(-1) ((1-sqrt(65))/4) + 180^@ ~~ 330.473^@# Note that your calculator may give a negative value for #cot^(-1) ((1-sqrt(65))/4)# (namely #~~ -29.527^@#). If it does then just add #180^@#. If your calculator does not have #cot^(-1)# then take the reciprocal and use #tan^(-1)#, but note that this will give values in the range #(-90^@, 90^@)#, so you will need to add an initial #180^@# to the #~~ -29.527^@# value to get a solution in the requested range.