Solve in the interval #0^@#, #360^@# #2csc^2x + cotx = 10# How?
1 Answer
Answer
Answer:
Explanation
Explanation:
Answer:
#cot^(1) ((1+sqrt(65))/4) ~~ 23.816^@#
#cot^(1) ((1sqrt(65))/4) ~~ 150.473^@#
#cot^(1) ((1+sqrt(65))/4) + 180^@ ~~ 203.816^@#
#cot^(1) ((1sqrt(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
#2cot^2 x + cot x  8 = 0#
This is in the form:
#at^2+bt+c = 0#
with
So using the quadratic formula, we have:
#cot x = (b+sqrt(b^24ac))/(2a)#
#color(white)(cot x) = (color(blue)(1)+sqrt(color(blue)(1)^24(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
#cot^(1) ((1+sqrt(65))/4) ~~ 23.816^@#
#cot^(1) ((1sqrt(65))/4) ~~ 150.473^@#
#cot^(1) ((1+sqrt(65))/4) + 180^@ ~~ 203.816^@#
#cot^(1) ((1sqrt(65))/4) + 180^@ ~~ 330.473^@#
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