Question 876351
{{{cos(z)=0}}} when {{{z=(1/2)pi}}} and {{{z=(3/2)pi}}}
So then,
{{{2x-pi=(1/2)pi}}} and {{{2x-pi=(3/2)pi}}}
{{{2x=(3/2)pi}}} and {{{2x=(5/2)pi}}}
{{{x=(3/4)pi}}} and {{{x=(5/4)pi}}}
And since cosine is periodic in {{{2pi}}}
{{{x=(3/4)pi +- 2pi*n}}} and {{{x=(5/4)pi +- 2pi*n}}}
You can combine those two together to get the Wolfram answer.