Solve in the interval #0^@#, #360^@# #2csc^2x + cotx = 10# How?

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Oct 24, 2017

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^@#

Explanation:

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.

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