Question

So I have a solving trig equations review, and this specific problem is giving me some troubles;

5-(3/5)cot=(25+root3)/5
It's supposed to be solved for what radians (pi/3, pi/6, etc) between 0 and 2pi. Any help would be gratefully recieved.

Answer 1

Answer: The equation is 5 - (3/5)cot(x) = (25 + √3)/5.

First, we can simplify the right-hand side by dividing both sides by 5:

(25 + √3)/5 = 5 + √3/5

Next, we can use the identity cot(x) = 1/tan(x) to rewrite the left-hand side:

5 - (3/5)cot(x) = 5 - (3/5)(1/tan(x)) = 5 - 3tan(x)/5

Now we have the equation 5 - 3tan(x)/5 = 5 + √3/5.

Subtracting 5 from both sides, we get:

-3tan(x)/5 = √3/5

Multiplying both sides by -5/3, we get:

tan(x) = -√3/3

Taking the arctangent of both sides, we get:

x = arctan(-√3/3)

Since the range of arctan is (-π/2, π/2), we need to add π to get the other solutions:

x = arctan(-√3/3) + π

x = arctan(-√3/3) + 2π

Using a calculator, we find that arctan(-√3/3) is approximately -0.5236 radians, so the solutions are:

x ≈ 2.6179 radians, 5.7596 radians, 8.9013 radians

Since we are looking for solutions between 0 and 2π, we can add or subtract multiples of 2π to get:

x ≈ 2.6179 radians, 5.7596 radians, 8.9013 radians, 11.0430 radians, 14.1847 radians, 17.3264 radians

These are the solutions to the equation 5 - (3/5)cot(x) = (25 + √3)/5 between 0 and 2π.

Explanation: