
How do you find possible values of x for $\sec x = \sqrt 2 $.
Answer
491.7k+ views
Hint: In the given question, we are required to find all the possible values of $\theta $ that satisfy the given trigonometric equation $\sec x = \sqrt 2 $. For solving such type of questions where we have to solve trigonometric equations, we need to have basic knowledge of algebraic rules and identities as well as a strong grip on trigonometric formulae and identities.
Complete step by step solution:
We have to solve the given trigonometric equation $\sec x = \sqrt 2 $ . We know that $\sec \left( \theta \right) = \dfrac{1}{{\cos \left( \theta \right)}}$ .
So, converting $\sec \theta $ into $\cos \theta $ using the basic trigonometric formula, we get,
$ \Rightarrow \dfrac{1}{{\cos x}} = \sqrt 2 $
Now, Shifting the $\cos x$ to the right hand side of the equation and isolating the trigonometric ratio, we get,
$ \Rightarrow \cos x = \dfrac{1}{{\sqrt 2 }}$
So, the above trigonometric equation is the simplified form of the one given in the question. Now, we have to look for values of x such that $\cos x = \dfrac{1}{{\sqrt 2 }}$ . So, we need a general solution for all the values of x for which $\cos x = \dfrac{1}{{\sqrt 2 }}$ .
We know that the general solution for the equation $\cos \left( \theta \right) = \cos \left( \phi \right)$ is $\theta = 2n\pi \pm \phi $. So, first we have to convert $\cos x = \dfrac{1}{{\sqrt 2 }}$ into $\cos \left( \theta \right) = \cos \left( \phi \right)$ form. So, we get,
$\cos x = \cos \left( {\dfrac{\pi }{4}} \right)$
Hence, for $\cos x = \cos \left( {\dfrac{\pi }{4}} \right)$ , we have \[x = 2n\pi \pm \dfrac{\pi }{4}\] .
Therefore, the possible values of x for $\sec x = \sqrt 2 $ are \[x = 2n\pi \pm \dfrac{\pi }{4}\] where n is any integer.
Note:
The general solution of a given trigonometric solution may differ in form, but actually represents the correct solution. The different forms of general equations are interconvertible into each other. Secant and cosine are reciprocal trigonometric functions. We must know the values of the trigonometric functions such as sine, cosine, etc. for some standard angles like $\dfrac{\pi }{6},\dfrac{\pi }{4},\dfrac{\pi }{3}$.
Complete step by step solution:
We have to solve the given trigonometric equation $\sec x = \sqrt 2 $ . We know that $\sec \left( \theta \right) = \dfrac{1}{{\cos \left( \theta \right)}}$ .
So, converting $\sec \theta $ into $\cos \theta $ using the basic trigonometric formula, we get,
$ \Rightarrow \dfrac{1}{{\cos x}} = \sqrt 2 $
Now, Shifting the $\cos x$ to the right hand side of the equation and isolating the trigonometric ratio, we get,
$ \Rightarrow \cos x = \dfrac{1}{{\sqrt 2 }}$
So, the above trigonometric equation is the simplified form of the one given in the question. Now, we have to look for values of x such that $\cos x = \dfrac{1}{{\sqrt 2 }}$ . So, we need a general solution for all the values of x for which $\cos x = \dfrac{1}{{\sqrt 2 }}$ .
We know that the general solution for the equation $\cos \left( \theta \right) = \cos \left( \phi \right)$ is $\theta = 2n\pi \pm \phi $. So, first we have to convert $\cos x = \dfrac{1}{{\sqrt 2 }}$ into $\cos \left( \theta \right) = \cos \left( \phi \right)$ form. So, we get,
$\cos x = \cos \left( {\dfrac{\pi }{4}} \right)$
Hence, for $\cos x = \cos \left( {\dfrac{\pi }{4}} \right)$ , we have \[x = 2n\pi \pm \dfrac{\pi }{4}\] .
Therefore, the possible values of x for $\sec x = \sqrt 2 $ are \[x = 2n\pi \pm \dfrac{\pi }{4}\] where n is any integer.
Note:
The general solution of a given trigonometric solution may differ in form, but actually represents the correct solution. The different forms of general equations are interconvertible into each other. Secant and cosine are reciprocal trigonometric functions. We must know the values of the trigonometric functions such as sine, cosine, etc. for some standard angles like $\dfrac{\pi }{6},\dfrac{\pi }{4},\dfrac{\pi }{3}$.
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