1)2cos(π/4-3x)=√2 2)tg (π/4-x/2)=-1 Помогите решить

1)cos(π/4-3x)=√2/2 π/4-3x=+-arccos(√2/2)+2[latex] \pi [/latex]n π/4-3x=+-[latex] \pi [/latex]/4+2[latex] \pi [/latex]n -3x=+-[latex] \pi [/latex]/4+2[latex] \pi [/latex]n-[latex] \pi [/latex]/4 x=+-(-[latex] \pi [/latex]/12)-2[latex] \pi [/latex]n/3+[latex] \pi [/latex]/12 2)π/4-x/2=arctg(-1)+[latex] \pi [/latex]n π/4-x/2=-[latex] \pi [/latex]/4+[latex] \pi [/latex]n -x/2=-[latex] \pi [/latex]/4+[latex] \pi [/latex]n-[latex] \pi [/latex]/4 x=[latex] \pi [/latex]/8-[latex] \pi [/latex]n/2+[latex] \pi [/latex]/8