Sin^4x-cos^4x=-sin^4x решите уравнение

Решите задачу:

$sin^4x + sin^4x = cos^4x \\ \\ 2sin^4x = cos^4x \ \ \ \ \ \ \ \ \ |:cos^4x \\ \\ 2tg^4x = 1 \\ \\ tg^4x = \dfrac{1}{2} \\ \\ 1) \ tgx = \dfrac{1}{ \sqrt[4]{2} } \\ \\ \boxed{x = arctg \dfrac{1}{ \sqrt[4]{2} } + \pi n, \ n \in Z} \\ \\ 2) \ tgx = -\dfrac{1}{ \sqrt[4]{2} } \\ \\ \boxed{x = arctg \bigg (-\dfrac{1}{ \sqrt[4]{2} } \bigg ) + \pi k, \ k \in Z }$