Question 961317
for: [0, 2π)
(square root of 2)(sinx)(cosx)+(cosx)=0
{{{sqrt(2)sinxcosx+cosx=0}}}
{{{cosx(sqrt(2)sinx+1)=0}}}
cosx=0
x=π/2, 3π/2
or
√2sinx+1=0
sinx=-1/√2=-√2/2
x=5π/4, 7π/4