Calculating $ x= \sqrt{4+\sqrt{4-\sqrt{ 4+\dots}}} $ If $ x= \sqrt{4+\sqrt{4-\sqrt{ 4+\sqrt{4-\sqrt{ 4+\sqrt{4-\dots}}}}}} $
then find value of 2x-1
I tried the usual strategy of squaring and substituting the rest of series by x again but could not solve.

Since $$x=\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-\cdots}}}},$$ you have $$x^2=4+\sqrt{4-\sqrt{4+\sqrt{4-\cdots}}},$$ so $$(x^2-4)^2=4-\sqrt{4+\sqrt{4-\cdots}}.$$ Hence, $$(x^2-4)^2=4-x,$$ which you can try to solve.
