Question

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

Answer 1

`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`