How to find $ \tan \left(\frac{x}{2}\right) $ knowing that $\cos \left(x\right)+\sin \left(x\right)=\frac{7}{5} $ Good evening to everyone. I don't know how to find $ \tan \left(\frac{x}{2}\right) $ knowing that $$\cos \left(x\right)+\sin \left(x\right)=\frac{7}{5} $$ and x$\in (0,\frac{\pi}{3})$ Here's what I've tried:
$$\tan \left(\frac{x}{2}\right) = \frac{1-\cos \left(x\right)}{\frac{7}{5}- \cos \left(x\right)}$$ But I don't know what to do from here. Can someone explain to me how to solve this? Thanks for any answers.

Since $$\sin { x } =2\sin { \frac { x }{ 2 }  } \cos { \frac { x }{ 2 }  } \\ \cos { x } =\cos ^{ 2 }{ \frac { x }{ 2 }  } -\sin ^{ 2 }{ \frac { x }{ 2 }  } $$
so 

$$\cos  \left( x \right) +\sin  \left( x \right) =\frac { 7 }{ 5 } \\ 5\left( \cos ^{ 2 }{ \frac { x }{ 2 } -\sin ^{ 2 }{ \frac { x }{ 2 }  }  }  \right) +10\sin { \frac { x }{ 2 }  } \cos { \frac { x }{ 2 }  } =7\left( \cos ^{ 2 }{ \frac { x }{ 2 } +\sin ^{ 2 }{ \frac { x }{ 2 }  }  }  \right) \\ 12\sin ^{ 2 }{ \frac { x }{ 2 } - } 10\sin { \frac { x }{ 2 }  } \cos { \frac { x }{ 2 }  } +2\cos ^{ 2 }{ \frac { x }{ 2 } =0 } \\ 6\tan ^{ 2 }{ \frac { x }{ 2 } -5\tan { \frac { x }{ 2 } +1 } =0 } \\ \tan { \frac { x }{ 2 } =\frac { 5\pm 1 }{ 12 }  }  $$

