Question

How do you find the general solutions for `3 (tan A + sec A) = 2cot A`?

Answer 1

Solve 3(tan A + sec A) = 2cot A

Ans: 270; 23.58 and 156.42 deg

Explanation:

`3(sin A/cos A + 1/cos A) - (2cos A)/sin A = 0`
`(3(sin A + 1))/cos A - (2cos A)/sin A = 0`
`((3sin A + 3)(sin A))/(sin A.cos A) - (2cos^2 A)/(sin A.cos A) = 0`

Develop the numerator, and replace `cos^2 A` by `(1 - sin^2 A)`:
`3sin^2 A + 3sin A - 2(1 - sin^2 A) =`
`= 5sin^2 A + 3sin A - 2 = 0` (Conditions: sin A not 0, and sin B not 0)
Since a - b + c = 0 --> 2 real roots:
`sin A = -1` and `sin A = -c/a = 2/5`
a. sin A = - 1 --> `A = (3pi)/2` (or 290 deg)
b. `sin A = 2/5 = 0.4` --> `A = 23.58` deg and `A = 180 - 23.58 = 156.42` deg