Question

1) 1-корень из 2 * косинус а 2) tgа+корень из 3

Answer 1

1)1- $\sqrt{2}$cosa = 0
    - $\sqrt{2}$cosa = -1
    cosa =  $\frac{ \sqrt{2} }{2}$
    a = +-arccos $\frac{ \sqrt{2} }{2}$ + 2πk, k∈z
    a = +- $\frac{ \pi }{4}$ + 2πk, k ∈ z
2)tga +  $\sqrt{3}$ = 0
   tga = -  $\sqrt{3}$
   a = -arctg $\sqrt{3}$ + πk, k ∈ z
   a = - $\frac{ \pi}{3}$ + πk, k ∈ z

Answer 2

I'm sorry, but I'm not exactly sure what you're asking for. It seems like you have provided two mathematical expressions, but there is no question or context to guide my response.

Can you please provide more information or clarify your question so that I can better assist you?