Find all solutions of the equation..

[LEG7930]

sec(x)-tan(x)=cos(x)
I know that there are 2 answers. They need to be on the interval of 0 to 2pi.

What I tried was the following:

sec(x)=1/cos
tan(x)=sin(x)/cos(x)
so
(1/cos(x))-(sin(x)/cos(x))=cos(x)
Then I found a common denominator...
(1-sin(x))/cos(x)=(cos^2(x)/cos(x))
Then I multiplied by the reciprocal of (cos^2(x)/cos(x))
(1-sin(x))/cos(x)(cos(x)/cos^2(x))=(cos^2(x)/cos(x))(cos(x)/cos^2(x))
so that simplifies to...
(1-sin(x))/(cos^2(x))=1
and now I am stuck. I don't even know if I was doing it correctly. Please HELP!

 

[soroban]

Hello, LEG7930!

\(\displaystyle \text{Solve: }\:\sec x -\tan x \:=\:\cos x \quad \text{ on }[0,2\pi)\)

\(\displaystyle \text{Note that: }\,x \,\ne\,\tfrac{\pi}{2},\:\tfrac{3\pi}{2} \quad\Rightarrow\quad \cos x \,\ne\,0\)


\(\displaystyle \text{We have: }\:\frac{1}{\cos x} - \frac{\sin x}{\cos x} \:=\:\cos x\)

\(\displaystyle \text{Multiply by }\cos x\!:\;\;1 - \sin x \:=\:\cos^2\!x \quad\Rightarrow\quad 1 - \sin x \:=\:1-\sin^2\!x\)

. . . . . . . . . . . . . .\(\displaystyle \sin^2\!x - \sin x \:=\:0 \quad\Rightarrow\quad \sin x(\sin x - 1) \:=\:0\)


. . . . . . . . \(\displaystyle \sin x \:=\: 0 \quad\Rightarrow\quad \boxed{x \:=\: 0,\,\pi}\)
We have:
. . . . . . . . \(\displaystyle \sin x - 1 \:=\: 0 \quad\Rightarrow\quad \sin x \:=\:1 \quad\Rightarrow\quad x \:=\:\tfrac{\pi}{2}\quad\text{(not allowed)}\)

 

[LEG7930]

thanks for your quick response. Its funny I had actually gotten those answers but i needed to put them in least to greatest (so 0 first then pi) and i hadnt done that lol

thanks!!!