Solve the equation for solutions over the interval [0,360) degrees.

Question

Cot theta + 4 csc theta = 5

what is the solution set { }

Answer 1

Perhaps there's an easier way but I couldn't think of it if there is.

Using the Pythagorean Identity csc²(θ) = cot²(θ) + 1 and solving for csc(θ) by taking the square root of both sides, we get:

csc(θ) = √(cot²(θ) + 1)

Substituting into the original equation, cot(θ) + 4csc(θ) = 5, we get:

cot(θ) + 4√(cot²(θ) + 1) = 5

Subtracting cot(θ) from both sides we get:

4√(cot²(θ) + 1) = 5 - cot(θ)

Squaring both sides, we get:

16(cot²(θ) + 1) = 25 - 10cot(θ) + cot²(θ) or:

16cot²(θ) + 16 = 25 - 10cot(θ) + cot²(θ) or

15cot²(θ) + 10cot(θ) - 9 = 0

Using the substitution x = cot(θ) to make the quadratic formula more "visible" we get:

15x² + 10x - 9 = 0

Using the quadratic formula:

x = [-10 ± √(10² - 4(15)(-9))]/((2)(15)

x = (-10 ± √640)/30

x = 0.5099407098

x = -1.176607376

Since x = cot(θ) then 1/x = tan(θ) so:

θ = tan^-1(1/0.5099407098) = 1.099227812 radians

but tan(θ) is positive in Q1 and Q3 the Q3 angle would be the Q1 angle plus π which would be 4.240820466 however when we try both of these in the original equation, only 1.099227812 works so the other is an extraneous solution.

For the other zero, θ = tan^-1(1/-1.176607376) = -0.7044366927 radians but to get a positive we add 2π and the result is 5.578748614. Again, this time we have to realize tangent is negative in Q4 and Q2. To get the Q2 angle, we subtract π so 5.578748614 - π = 2.43715596. This time, when we try the answers only 2.43715596 works.

So the answers (in radian measure) are θ = 1.099227812 and θ = 2.43715596. To turn these into degrees, we use the conversion 180° = π radians so θ = 62.98111435° and θ = 139.6387505°