Given interval is `[0,2pi]` .
And given identity is `secx+tanx=1`
which we can write as `1/(cosx)+(sinx)/(cosx)=1`
`or` , `(1+sinx)/(cosx)=1`
or, `(1+sinx)=cosx`
or, `1+sinx-cosx=0`
or, `1+2sin(x/2)cos(x/2)-(1-2sin^2(x/2))=0`
or, `1+2sin(x/2)cos(x/2)-1+2sin^2(x/2)=0`
or, `2sin(x/2)cos(x/2)+2sin^2(x/2)=0`
or, `2sin(x/2)[cos(x/2)+sin(x/2)]=0`
`or,` `2sin(x/2)=0` and `cos(x/2)+sin(x/2)=0` .
Now, `2sin(x/2)=0` `rArrsin(x/2)=0` `rArrsin(x/2)=sin(npi)` , n=0,1,2. As per the need of our
problem.
`rArr x/2=npi` or, `x=2npi` for n=0,1 as our interval is `[0,2pi]` .
So, `x=0` or, `x=2pi` .
Now, if `cos(x/2)+sin(x/2)=0` `rArr1+(sin(x/2))/cos(x/2)=0`
`rArrtan(x/2)=-1` or, `tan(x/2)=tan(pi/2+pi/4)`
or, `x/2=(pi/2+pi/4)`
or, `x=pi+pi/2`
or, `x=3pi/2` .
Clearly, `x=0` and `x=2pi` are the solutions satisfying our given identity over the given interval.
Solution
Given interval is `[0,2pi]` .
And given identity is `secx+tanx=1`
which we can write as `1/(cosx)+(sinx)/(cosx)=1`
`or` , `(1+sinx)/(cosx)=1`
or, `(1+sinx)=cosx`
or, `1+sinx-cosx=0`
or, `1+2sin(x/2)cos(x/2)-(1-2sin^2(x/2))=0`
or, `1+2sin(x/2)cos(x/2)-1+2sin^2(x/2)=0`
or, `2sin(x/2)cos(x/2)+2sin^2(x/2)=0`
or, `2sin(x/2)[cos(x/2)+sin(x/2)]=0`
`or,` `2sin(x/2)=0` and `cos(x/2)+sin(x/2)=0` .
Now, `2sin(x/2)=0` `rArrsin(x/2)=0` `rArrsin(x/2)=sin(npi)` , n=0,1,2. As per the need of our
problem.
`rArr x/2=npi` or, `x=2npi` for n=0,1 as our interval is `[0,2pi]` .
So, `x=0` or, `x=2pi` .
Now, if `cos(x/2)+sin(x/2)=0` `rArr1+(sin(x/2))/cos(x/2)=0`
`rArrtan(x/2)=-1` or, `tan(x/2)=tan(pi/2+pi/4)`
or, `x/2=(pi/2+pi/4)`
or, `x=pi+pi/2`
or, `x=3pi/2` .
Clearly, `x=0` and `x=2pi` are the solutions satisfying our given identity over the given interval..