Question

How do you solve $\sec x = 1$?

Answer 1

Hint:The secant is the reciprocal of cosine and represents the ratio of hypotenuse and adjacent side in a right-angled triangle. Use the equation $\sec x = \dfrac{1}{{\cos x}} = \dfrac{{Hypotenuse}}{{Adjacent}}$ and $\sec x = 1$ can be combined to find the value of cosine. Then after getting the equation in the form of cosine, you can use the solutions of cosine to get the solution.

Complete step by step answer:
Here in this problem, we have given an expression secant trigonometric ratio, i.e. $\sec x = 1$. And we need to find the solution to this problem.
Before starting with the solution to this problem, we must understand a few concepts regarding the trigonometric ratios and secant. In a right triangle, the secant of an angle is the length of the hypotenuse divided by the length of the adjacent side. In a formula, it is abbreviated to just 'sec'.
Of the six possible trigonometric functions, secant, cotangent, and cosecant, are rarely used.
Secant can be represented as the reciprocal of the cosine:
$ \Rightarrow \sec x = \dfrac{1}{{\cos x}} = \dfrac{{Hypotenuse}}{{Adjacent}}$.

So here we have:
$ \Rightarrow \sec x = 1$ , and we need to find the unknown angle $x$
According to the definition of secant, we can write it as:
$ \Rightarrow \sec x = \dfrac{1}{{\cos x}} = 1$
Therefore, we get the value for the cosine as:
$ \Rightarrow \sec x = \dfrac{1}{{\cos x}} = 1 \Rightarrow \cos x = 1$
The above expression represents that the cosine has its value $1$ at the angle $'x'$
As we know that the cosine takes the value $1$ for the integral multiples of $2\pi $, i.e. $2n\pi $, for all the $'n'$ integers.
Therefore, the values for $'x'$ can be given by:
$ \Rightarrow x = 0,2\pi , - 2\pi ,4\pi , - 4\pi {\text{ etc}}$
The above values can also be represented as:
$ \Rightarrow x = 2n\pi $ , where ‘n’ is an integer.

Note: In this problem, we used the solution for $\cos x = 1$ to find the required answer. An alternative approach can be to use the definition of cosine to find the solution for $\cos x = 1$ , i.e. $\cos x = \dfrac{{Adjacent}}{{Hypotenuse}} = 1 \Rightarrow Hypotenuse = Adjacent{\text{ }}Side$ . But in a right-angled triangle, hypotenuse and adjacent side can be equal when the angle is zero, or they both overlap each other.