Question

How do you solve \[\sec x=\tan x+1\] for \[0\le x\le 2\pi \] ?

Answer 1

Hint: In the given solution, you first need to change the equation in sine and cosine form and after solving the equation further try to make the equation that is comparable to the compound trigonometric identities and you need to divide the converted equation and then you need to apply compound trigonometric identity to get the solution by considering the range of the trigonometric function.

Formula used:
\[\sec x=\dfrac{1}{\cos x}\]
\[\Rightarrow\tan x=\dfrac{\sin x}{\cos x}\]
To make the equation comparable to compound trigonometric formula, we will divide the resultant equation by
\[\sqrt{{{\left( coefficient\ of\ \sin x \right)}^{2}}+{{\left( coefficient\ of\ \cos x \right)}^{2}}}\]
To find the value of \[x\]:
\[x=2n\pi \pm \dfrac{\pi }{4}+\dfrac{\pi }{4},where\,n\in I\]
\[\Rightarrow\cos x\cos y+\sin x\sin y=\cos \left( x+y \right)\]

Complete step by step solution:
We have the given equation:
\[\sec x=\tan x+1\]
With the help of trigonometric formulas, we know the value of,
\[\sec x=\dfrac{1}{\cos x}\]
\[\Rightarrow\tan x=\dfrac{\sin x}{\cos x}\]
Therefore, by substituting the value of \[\sec x\] and \[\tan x\] in the given equation,
\[\dfrac{1}{\cos x}=\dfrac{\sin x}{\cos x}+1\]
Simplifying the resultant equation, we obtain
\[\dfrac{1}{\cos x}=\dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\cos x}\]
\[\Rightarrow\dfrac{1}{\cos x}=\dfrac{\sin x+\cos x}{\cos x}\]

By transposing \[\cos x\] from the denomination of the RHS to the LHS of equals to sign,
There will be the inverse of mathematical operation, on transposing to other side division will change to multiplication,
Simplifying further, we get
\[1=\sin x+\cos x\]
\[\Rightarrow\cos x+\sin x=1\]
Now, to make the equation comparable to compound trigonometric formula, we will divide the resultant equation by
\[\sqrt{{{\left( coefficient\ of\ \sin x \right)}^{2}}+{{\left( coefficient\ of\ \cos x \right)}^{2}}}\]
\[\Rightarrow\cos x+\sin x=1\]
\[\Rightarrow\dfrac{\cos x+\sin x}{\sqrt{{{1}^{2}}+{{1}^{2}}}}=\dfrac{1}{\sqrt{{{1}^{2}}+{{1}^{2}}}}\]
\[\Rightarrow\dfrac{\cos x+\sin x}{\sqrt{2}}=\dfrac{1}{\sqrt{2}}\]
\[\Rightarrow\dfrac{\cos x}{\sqrt{2}}+\dfrac{\sin x}{\sqrt{2}}=\dfrac{1}{\sqrt{2}}\]

As we all know that,
\[\cos \dfrac{\pi }{4}=\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\]
Replace \[\dfrac{1}{\sqrt{2}}\] with \[\cos \dfrac{\pi }{4}\ and\ \sin \dfrac{\pi }{4}\], we get
\[\cos x\cos \dfrac{\pi }{4}+\sin x\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\]
By remembering the compound angle formula of trigonometric identities, i.e.
\[\cos x\cos y+\sin x\sin y=\cos \left( x+y \right)\]
By using this formula, we get
\[\cos x\cos \dfrac{\pi }{4}+\sin x\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\]
\[\Rightarrow\cos \left( x-\dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}\]

By doing general solution,
\[\cos \theta =\dfrac{1}{\sqrt{2}}\],
Which is,
 \[x-\dfrac{\pi }{4}=2n\pi \pm \dfrac{\pi }{4},where\,n\in I\]
Transposing \[\dfrac{\pi }{4}\]to the other side, we get
\[x=2n\pi \pm \dfrac{\pi }{4}+\dfrac{\pi }{4},where\,n\in I\]
Checking for the value of \[x=2n\pi +\dfrac{\pi }{4}+\dfrac{\pi }{4}=2n\pi +\dfrac{\pi }{2}\], here \[\sec x\] and \[\tan x\] are undefined.
Checking for the value of \[x=2n\pi -\dfrac{\pi }{4}+\dfrac{\pi }{4}=2n\pi \], here \[\sec x=1\] and \[\tan x=0\], which is satisfying the given range of the equation i.e. \[\sec x=\tan x+1\] for \[0\le x\le 2\pi \].

Therefore, the required solution is \[x=2n\pi \].

Note: While evaluating the trigonometric angles, it is always useful to bring the terms in the form of known quantities. Thus, it is always better to convert \[\sec x,\cos ecx\ and\ \cot x\] in the terms of \[\cos x,\sin x\ and\ \tan x\]. It is always better to remember some basic trigonometric identities, which makes the question to solve easier.