Question

How do you solve $ \tan x+\sec x=1 $ ? \[\]

Answer 1

Hint: We convert the tangent and secant of the equation into sine and cosine and then simplify. We shall obtain the equation $ \cos x-\sin x=1 $ where we square both sides to have $ \sin x\cos x=0 $ . We use the general solution of $ \sin x=0 $ that is $ x=n\pi ,n\in Z $ and the general solution of $ \cos x=0 $ that is $ x=\left( 2n+1 \right)\dfrac{\pi }{2} $ which we reject because $ \tan x,\sec x $ are undefined for $ \cos x=0 $ . \[\]

Complete step by step answer:
The general solution of $ \sin x=0 $ with principal solution $ \theta =0 $ are given by $ x=n\pi $ and the general solution of $ \cos x=0 $ with principal solution $ \theta =\dfrac{\pi }{2} $ are given by $ x=\left( 2n+1 \right)\dfrac{\pi }{2} $ .Here $ n $ is an arbitrary integer of our choice. \[\]
We are asked to solve the following trigonometric equation
 $ \tan x+\sec x=1 $ \[\]
Let us convert the above equation into sine and cosine using basic trigonometric ratios $ \tan \theta =\dfrac{\sin \theta }{\cos \theta },\sec \theta =\dfrac{1}{\cos \theta } $ to have;
 $ \begin{align}
  & \dfrac{\sin x}{\cos x}+\dfrac{1}{\cos x}=1 \\
 & \Rightarrow \dfrac{\sin x+1}{\cos x}=1 \\
\end{align} $
We multiply both sides of above equation by $ \cos x $ to have;
\[\begin{align}
  & \Rightarrow \sin x+1=\cos x \\
 & \Rightarrow \cos x-\sin x=1 \\
\end{align}\]
We square both sides of above equation to have;
\[\begin{align}
  & \Rightarrow {{\left( \cos -\sin x \right)}^{2}}={{1}^{2}} \\
 & \Rightarrow {{\cos }^{2}}x+{{\sin }^{2}}x-2\sin \cos x=1 \\
\end{align}\]
We use the Pythagorean trigonometric identity $ {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 $ in the above step to have;
\[\begin{align}
  & \Rightarrow 1-2\sin x\cos x=1 \\
 & \Rightarrow -2\sin x\cos x=0 \\
 & \Rightarrow \sin x\cos x=0 \\
\end{align}\]
So we have two cases here $ \sin x=0 $ and $ \cos x=0 $ . We reject $ \cos x=0 $ because $ \tan x,\sec x $ are not defined for $ \cos x=0 $ which means at $ x=\left( 2n+1 \right)\dfrac{\pi }{2} $ . So the possible solutions of the given equation are also the solutions of $ \sin x=0 $ which are
\[x=n\pi =...,-2\pi ,-2\pi ,0,\pi ,2\pi ,...\]
When we put the above values $ x=n\pi $ in original equation $ \tan x+\sec x=1 $ , it only satisfies for even $ n $ and for odd $ n $ the solutions become extraneous. So the solutions are
\[x=...,-2\pi ,0,2\pi ,...\]

Note:
 We note that a trigonometric equation is an equation with trigonometric functions with unknown arguments as a measure of angles. When we are asked to solve a trigonometric equation we have to find all possible measures of unknown angles. We know that the first solution of the trigonometric equation within the interval $ \left[ 0,2\pi \right] $ is called principal solution and using periodicity all possible solutions obtained with integer $ n $ are called general solutions.