Infinitely repeated game: prisoners' dilemma
the payoff from play the trigger strategy will be:
$$ \sum_{i=0}^{\infty}(x+1) \delta^i=\frac{x+1}{1-\delta}$$ 

if I deviate and I play $l$ or $d$ the payoff will be
$$ 2x + \sum_{i=1}^{\infty}x \delta^i = 2x+ x\frac{\delta}{1-\delta}=\frac{2x(1-\delta) + x\delta}{1-\delta}= \frac{x(2-\delta)}{1-\delta}$$

then, the condition is $$ \frac{x+1}{1-\delta} \geq \frac{x(2-\delta)}{1-\delta}$$

I hope that this will help you ;)