Question

The general solution of the equation \(\tan x+\sec x=2\,\cos x\) is

• A. \(x=n\pi +(-1)^{n}\left(\frac{\pi }{3}\right),n\in Z\)
• B. \(x=n\pi +(-1)^{n+1}\left(\frac{\pi }{2}\right),n\in Z\)
• C. \(x=n\pi +(-1)^{n}\left(\frac{\pi }{4}\right),n\in Z\)
• D. \(x=n\pi +(-1)^{n}\left(\frac{\pi }{6}\right),n\in Z\)

Hint:

Convert to sine and cosine and form a quadratic in sin x, then discard the value that makes cos x zero.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Write \(\tan x+\sec x=\dfrac{\sin x+1}{\cos x}\).

Step 2: Form an equation
\(\dfrac{1+\sin x}{\cos x}=2\cos x\Rightarrow 1+\sin x=2\cos^{2}x=2(1-\sin^{2}x)\).

Step 3: Quadratic in sin x
\(2\sin^{2}x+\sin x-1=0\Rightarrow(2\sin x-1)(\sin x+1)=0\).
So \(\sin x=\dfrac{1}{2}\) or \(\sin x=-1\).

Step 4: Reject invalid root
\(\sin x=-1\) makes \(\cos x=0\), where \(\tan x\) is undefined, so reject it.

Step 5: General solution
\(\sin x=\sin\dfrac{\pi}{6}\Rightarrow x=n\pi+(-1)^{n}\dfrac{\pi}{6}\).

Final Answer:
The general solution is x = n pi + (-1)^n (pi/6). \[ \boxed{\text{(D) }x=n\pi +(-1)^{n}\left(\dfrac{\pi }{6}\right),n\in Z} \]