Ctgx + sinx/1+cosx =2

We start with the equation: \[ \frac{\cot x + \sin x}{1 + \cos x} = 2 \] First, we can rewrite \(\cot x\) in terms of sine and cosine: \[ \cot x = \frac{\cos x}{\sin x} \] Substituting this into the equation gives us: \[ \frac{\frac{\cos x}{\sin x} + \sin x}{1 + \cos x} = 2 \] To simplify the left side, we can combine the terms in the numerator: \[ \frac{\frac{\cos x + \sin^2 x}{\sin x}}{1 + \cos x} = 2 \] This can be rewritten as: \[ \frac{\cos x + \sin^2 x}{\sin x(1 + \cos x)} = 2 \] Now, we can cross-multiply to eliminate the fraction: \[ \cos x + \sin^2 x = 2\sin x(1 + \cos x) \] Distributing on the right side provides: \[ \cos x + \sin^2 x = 2\sin x + 2\sin x \cos x \] Now we can rearrange the equation: \[ \sin^2 x - 2\sin x - 2\sin x \cos x + \cos x = 0 \] This is a trigonometric equation that can be somewhat complicated to solve algebraically. However, we can use trigonometric identities and substitutions. Recall that \(\sin^2 x = 1 - \cos^2 x\). Substituting gives us: \[ 1 - \cos^2 x - 2\sin x - 2\sin x \cos x + \cos x = 0 \] This still may not lead to simpler solutions. Instead, we could try specific values of \(x\) or manipulate the trigonometric functions further. By inspection or trying specific angles, it might be easier. For example, set \(x = \frac{\pi}{2}\), \(x = 0\), \(x = \frac{\pi}{4}\), etc., or use a numerical or graphical method. Another approach is to apply identities to analyze the possible solutions and their worthiness. However, the best analytical approach is to express everything based on a single trigonometric function (like \(\sin x\) or \(\cos x\)), or to manipulate it further until a solvable form is recognized. If you wish to proceed deeper this was the full transformation until a direct solvable method is used or numerical estimation helps lead a satisfactory conclusion. Would you like to try specific angle values, or should we continue manipulating the trigonometric identities further?