How do you solve 7=cot^2x+4cotx in the interval [0,360]?

1 Answer
Nghi N.
Oct 18, 2016

37^@15, 217^@15
10^@65, 190^@65

Explanation:

Bring the quadratic equation in cot x to standard form:
cot^2 x + 4cot x - 7 = 0
Use the new improved quadratic formula (Socratic Search)
D = d^2 = b^2 - 4ac = 16 + 28 = 44 --> d = +- 2sqrt11
There are 2 real root:
cot x = -b/(2a) +- d/(2a) = -4/2 +- 2sqrt11/2 = -2 +- sqrt11
cot x1 = -2 + sqrt11 = 1.32,
cot x2 = -2 - sqrt11 = - 5/32
Use calculator and unit circle -->
tan x1 = 1/1.32 = 0.7575 --> Two values of x1 -->
x1 = 37^@15, and x1 = 37.15 + 180 = 217^@15
tan x2 = 1/-5.32 = - 0.19 --> Two values of x2-->
x2 = 10^@65, and x2 = 10.65 + 180 = 190^@65

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