How do you solve $cosx/(1+sinx)+(1+sinx)/cosx=-4$ for $0<=x<=2pi$ ?

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Answer 1 · 2016-11-26T06:24:20

$x=(2pi)/3$ or $(4pi)/3$

We have $cosx/(1+sinx)+(1+sinx)/cosx=-4$ .

This can be written as

$(cos^2x+(1+sinx)^2)/(cosx(1+sinx))=-4$

or $(cos^2x+1+sin^2x+2sinx)/(cosx(1+sinx))=-4$

or $(1+1+2sinx)/(cosx(1+sinx))=-4$

or $(2(1+sinx))/(cosx(1+sinx))=-4$

or $2/cosx=-4$

or $cosx=-2/4=-1/2=cos(+-(2pi)/3)$

i.e. $x=2npi+-(2pi)/3$ , where $n$ is an integer

and if $0<=x<=2pi$ ,

$x=(2pi)/3$ or $(4pi)/3$

How do you solve cosx/(1+sinx)+(1+sinx)/cosx=-4cosx/(1+sinx)+(1+sinx)/cosx=-4 for 0<=x<=2pi0<=x<=2pi?