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Question

Question: How do you find possible values of x for \(\sec x = \sqrt 2 \)....

How do you find possible values of x for secx=2\sec x = \sqrt 2 .

Explanation

Solution

In the given question, we are required to find all the possible values of θ\theta that satisfy the given trigonometric equation secx=2\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 secx=2\sec x = \sqrt 2 . We know that sec(θ)=1cos(θ)\sec \left( \theta \right) = \dfrac{1}{{\cos \left( \theta \right)}} .
So, converting secθ\sec \theta into cosθ\cos \theta using the basic trigonometric formula, we get,
1cosx=2\Rightarrow \dfrac{1}{{\cos x}} = \sqrt 2
Now, Shifting the cosx\cos x to the right hand side of the equation and isolating the trigonometric ratio, we get,
cosx=12\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 cosx=12\cos x = \dfrac{1}{{\sqrt 2 }} . So, we need a general solution for all the values of x for which cosx=12\cos x = \dfrac{1}{{\sqrt 2 }} .
We know that the general solution for the equation cos(θ)=cos(ϕ)\cos \left( \theta \right) = \cos \left( \phi \right) is θ=2nπ±ϕ\theta = 2n\pi \pm \phi . So, first we have to convert cosx=12\cos x = \dfrac{1}{{\sqrt 2 }} into cos(θ)=cos(ϕ)\cos \left( \theta \right) = \cos \left( \phi \right) form. So, we get,
cosx=cos(π4)\cos x = \cos \left( {\dfrac{\pi }{4}} \right)
Hence, for cosx=cos(π4)\cos x = \cos \left( {\dfrac{\pi }{4}} \right) , we have x=2nπ±π4x = 2n\pi \pm \dfrac{\pi }{4} .
Therefore, the possible values of x for secx=2\sec x = \sqrt 2 are x=2nπ±π4x = 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 π6,π4,π3\dfrac{\pi }{6},\dfrac{\pi }{4},\dfrac{\pi }{3}.