Question

Tanx+secx=1... Find the solutions of the equation that are in the interval [0,2pi)

Answer 1

Use the fact that:
\[\tan x=\frac{\sin x}{\cos x}\]
and
\[\sec x=\frac{1}{\cos x}\]
We can change substitute the values on the RHS in your equation.
Your ORIGINAL equation was this:
\[\tan x +\sec x=1\]
Your ADJUSTED equation is now this when you substitute both tanx and secx with the values I've listed:
\[\frac{\sin x}{\cos x}+\frac{1}{\cos x}=1\]
Now you combine the two fractions together since both fractions have the same denominator:
\[\frac{\sin x+1}{\cos x}=1\]
Now we can move cosx to the RHS, to give you this:
\[\sin x+1=\cos x\]
Next step is to put sinx and cosx together on one side and the 1 on its own to the other side:
\[1=\cos x -\sin x\]
\[\cos x-\sin x =1\]
Now there is a method known as the Auxiliary angle method. You have to get the above equation in the form of this:
\[Rcos(x+\epsilon)=1\]
where: the auxiliary angle is the "e" symbol or commonly known as epsilon
\[R=\sqrt{a^2+b^2}\]
\[a\cos x\pm b\sin x=1\]
If you need help with the auxiliary angle method, it's best to ask your teacher when you get the time in class to do so.