How do you find the solution to #2cot^2theta-13cottheta+6=0# if #0<=theta<2pi#?

1 Answer
Nghi N
Dec 27, 2016

#9^@46, 63^@43, 189^@46, 243^@43#

Explanation:

Solve this quadratic equation for cot t
#f(t) = 2cot^2 t - 13cot t + 6 = 0#
#D = d^2 = b^2 - 4ac = 169 - 48 = 121# --> #d = +- 11#
There are 2 real roots:
#cot t = -b/(2a) +- d/(2a) = 13/4 +- 11/4 = (13 +- 11)/4#
#cot = 24/4 = 6# --> #tan t = 1/(cot t) = 1/6#
#cot t = 2/4 = 1/2# --> #tan t = 1/(cot t) = 2#
Use calculator and unit circle -->
a. #tan t = 1/6# --> #t = 9^@46# and #t = 180 + 9.46 = 189^@46#
b. tan t = 2 --># t = 63^@43# and #t = 180 + 63.43 = 243^@43#

Answers for (0, 360):
#9^@46, 63^@43, 189^@46, 243^@43#

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