Question 884078
solve: {{{2sin(2x)+cos(2x)+1=0}}} for {{{0<=x<=360}}}
***
{{{2sin(2x)+cos(2x)+1=0}}}
{{{cos(2x)+1=-2sin(2x)}}}
{{{cos(2x)+1=-2sqrt(1-cos^2(2x))}}}
square both sides
cos^2(2x)+2cos(2x)+1=4(1-cos^2(2x)
cos^2(2x)+2cos(2x)+1=4-4cos^2(2x)
5cos^2(2x)+2cos(2x)-3=0
(5cos(2x)-3)(cos(2x)+1)=0
..
(5cos(2x)-3)=0
cos(2x)=3/5
2x&#8776;53.13&#730; and 306.87
x&#8776;26.57&#730; and 153.43&#730;
reject, (extraneous roots) 
..
Solution:
cos(2x)+1=0
cos(2x)=-1
2x=180&#730;
x=90&#730;