The equation is:

cotg(X) = (sqrt(3)*cos(50º) + sin(50º))*(1-2*sqrt(3)*cos(50º)+2*sin(50º)) /

(sqrt(3)*sin(50º) - cos(50º))

Using a calculator, I find X = 50º.

How to develop the equation to find the solution?

Thanks.

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- Dec 21st 2011, 02:55 AMTOZEHow to solve an equation
The equation is:

cotg(X) = (sqrt(3)*cos(50º) + sin(50º))*(1-2*sqrt(3)*cos(50º)+2*sin(50º)) /

(sqrt(3)*sin(50º) - cos(50º))

Using a calculator, I find X = 50º.

How to develop the equation to find the solution?

Thanks. - Dec 21st 2011, 03:30 AMHallsofIvyRe: How to solve an equation
What do you mean by "develop" the equation? The right side is simply a number. Find that number and take its inverse cotangent.

- Dec 21st 2011, 06:28 AMTOZERe: How to solve an equation
Doing what you propose with the calculator we arrive to 50 degrees.

So, if X = 50, it would be possible to "develop" or simplify the trigonometric expression on the second member of the equation to arrive to cotg(50).

Don't you think so, considering that in the expression we have only trigonometric functions of the same angle? - Dec 21st 2011, 09:59 AMFernandoRevillaRe: How to solve an equation
Consider any angle $\displaystyle \alpha$ instead of $\displaystyle 50^0$ and expand $\displaystyle \cot \alpha$ in the following way:

$\displaystyle \cot \alpha =\cot \;[30^0+(\alpha -30^0)]=\frac{\cot 30^0\cdot \cot (\alpha -30^0)-1}{\cot 30^0+\cot (\alpha -30^0)}=$

$\displaystyle \frac{\sqrt{3}\cdot \dfrac{\cos (\alpha-30^0)}{\sin (\alpha-30^0)}-1}{\sqrt{3}+\dfrac{\cos (\alpha-30^0)}{\sin (\alpha-30^0)}}}=\ldots$

Let's see if you get the right side of the equation (with $\displaystyle \alpha$ instead of $\displaystyle 50^0$ ).

P.D. I haven't checked it, it is only a proposal.