Question

How do you solve `4cosx+3=2/cosx` in the interval `0<=x<=2pi`?

Answer 1

`63^@26; 296^@74`

Explanation:

Cross multiply, then bring equation to standard form:
`4cos^2 x + 3cosx - 2 = 0`
Solve this quadratic equation by the improved quadratic formula (Socratic Search)
`D = d^2 = b^2 - 4ac = 9 + 32 = 41` --> `d = +- sqrt41`
There are 2 real toots:
`cos x = - 3/8 +- sqrt41/8 = (-3 +- sqrt41)/8`
`cos x1 = (-3 + sqrt41)/8 = 0.45`
`cos x2 = (-3 - sqrt41)/8 = - 1.18` (Rejected as < -1)
Calculator and unit circle -->
cos x = 0.45 --> `x = +- 63^@26`
arc (- 63.26) is co-terminal to (360 - 63.26 = 296^@74)
Answers for (0, 360):
`63^@26; 296^@74`