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Question: If \[\sqrt 2 \sec x + \tan x = 1\], then the value of \[x\] is A.\[2n\pi + \dfrac{\pi }{3}\] B.\...

If 2secx+tanx=1\sqrt 2 \sec x + \tan x = 1, then the value of xx is
A.2nπ+π32n\pi + \dfrac{\pi }{3}
B.2nππ42n\pi - \dfrac{\pi }{4}
C.2nπ+π62n\pi + \dfrac{\pi }{6}
D.2nπ+π122n\pi + \dfrac{\pi }{{12}}

Explanation

Solution

Here we will use the trigonometric properties. First, we will convert the given equation in terms of the sin and cos function and simplify it. Then we will put the value of the constants in the equation in terms of the trigonometric function. We will them simplify the equation to get the value of xx.

Complete step-by-step answer:
Given equation is 2secx+tanx=1\sqrt 2 \sec x + \tan x = 1.
We will write the given equation in terms of the sine and cosine functions. We know that secant function is equal to the reciprocal of the cos function and tangent function is equal to the ratio of the sine to the cosine function. Therefore, we get
21cosx+sinxcosx=1\Rightarrow \sqrt 2 \dfrac{1}{{\cos x}} + \dfrac{{\sin x}}{{\cos x}} = 1
Now we will take the cos function common in the denominator and take it to the other side of the equation, we get
2+sinxcosx=1\Rightarrow \dfrac{{\sqrt 2 + \sin x}}{{\cos x}} = 1
2+sinx=cosx\Rightarrow \sqrt 2 + \sin x = \cos x
Now we will simplify the above equation, we get
cosxsinx=2\Rightarrow \cos x - \sin x = \sqrt 2
Dividing both sides by 2\sqrt 2 , we get
12cosx12sinx=1\Rightarrow \dfrac{1}{{\sqrt 2 }}\cos x - \dfrac{1}{{\sqrt 2 }}\sin x = 1
Now we will put the value of the constants in the equation in terms of the trigonometric functions.
We know that sinπ4=cosπ4=12\sin \dfrac{\pi }{4} = \cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }} and cos2nπ=1\cos 2n\pi = 1.
Now by putting these values in the above equation, we get
cosπ4cosxsinπ4sinx=cos2nπ\Rightarrow \cos \dfrac{\pi }{4}\cos x - \sin \dfrac{\pi }{4}\sin x = \cos 2n\pi
Now using the property of the trigonometry cos(A+B)=cosAcosBsinAsinB\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B, we get
cos(π4+x)=cos2nπ\Rightarrow \cos \left( {\dfrac{\pi }{4} + x} \right) = \cos 2n\pi
Now by cancelling the cos function from both sides of the equation, we get
π4+x=2nπ\Rightarrow \dfrac{\pi }{4} + x = 2n\pi
Now by solving this we will get the value of xx.
Subtracting π4\dfrac{\pi }{4} from both the sides, we get
x=2nππ4\Rightarrow x = 2n\pi - \dfrac{\pi }{4}
Hence the value of xx is 2nππ42n\pi - \dfrac{\pi }{4}.
So, option B is the correct option.

Note: We know that there are six basic trigonometric functions and they are sine, cosine, tangent, cosecant, secant and cotangent. Also, cosecant, secant and cotangent are reciprocal functions of sine, cosine and tangent respectively. While solving trigonometric equations we should always write the trigonometric functions in terms of sine and cosine function. This makes it easier for us to solve the given equation. Every trigonometric function is a periodic function that means that they repeat their value after a certain interval. These intervals are the multiples of 2π2\pi .