# Solve (1+cosx)tanx/2 = sinx

Question
Solve $$\displaystyle{\left({1}+{\cos{{x}}}\right)}\frac{{\tan{{x}}}}{{2}}={\sin{{x}}}$$

2020-12-31

$$\displaystyle{\cos{{\left({x}\right)}}}={2}{\left({\cos{{\left(\frac{{x}}{{2}}\right)}}}\right)}^{{2}}-{1},{s}{o}\ {\cos{{\left({x}\right)}}}+{1}={2}{\left({\cos{{\left(\frac{{x}}{{2}}\right)}}}\right)}^{{2}},$$
$$\displaystyle{\tan{{\left(\frac{{x}}{{2}}\right)}}}=\frac{{\sin{{\left(\frac{{x}}{{2}}\right)}}}}{{\cos{{\left(\frac{{x}}{{2}}\right)}}}},$$
$$\displaystyle{\sin{{\left({x}\right)}}}={2}{\sin{{\left(\frac{{x}}{{2}}\right)}}}{\cos{{\left(\frac{{x}}{{2}}\right)}}}.$$
Therefore, the expression on the left is: $$\displaystyle{2}{\sin{{\left(\frac{{x}}{{2}}\right)}}}{\cos{{\left(\frac{{x}}{{2}}\right)}}}={\sin{{\left({x}\right)}}}.$$

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