Question 945487
Solve the equation cotangent squared of the quantity of pi over 2 minus theta = secant of theta + 1 for for theta greater than or equal to 0 but less than 2 pi. 
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use x for theta
{{{cot^2(pi/2-x)=secx+1}}}
{{{tan^2(x)=1/cosx+1}}}
{{{sin^2(x)/cos^2(x)=1/cosx+1}}}
lcd: cos^2(x)
{{{sin^2(x)=cosx+cos^2(x)}}}
{{{1-cos^2(x)=cosx+cos^2(x)}}}
2cos^2(x)+cosx-1=0
(2cosx-1)(cosx+1)=0
cosx=1/2
x=π/3,5π/3
or
cosx=-1
x=π