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:
$$ \cos(x)=\frac{1-\tan^2\left(\frac{x}{2}\right)}{1+\tan^2\left(\frac{x}{2}\right)},\qquad\sin(x) =\frac{2\tan\left(\frac{x}{2}\right)}{1+\tan^2\left(\frac{x}{2}\right)}\tag{1}$$
we get:
$$ \left(1-\tan\left(\frac{x}{2}\right)\right)^2 = \frac{7}{5}\left(1+\tan^2\left(\frac{x}{2}\right)\right)\tag{2}$$
and $\tan\left(\frac{x}{2}\right)\color{red}{\in\left\{\frac{1}{2},\frac{1}{3}\right\}}$ can be found by solving a quadratic equation.
As an alternative, we may simply consider the (right) triangle with side lenghts $3,4,5$ and find its (unit) inradius.
