Question

The general solution of the equation \( \sqrt{6 - 5\cos x + 7\sin^2 x} - \cos x = 0 \) also satisfies the equation

• A. \( \tan x + \cot x = 2 \)
• B. \( \cot x + \cosec x = 1 \)
• C. \( \tan x + \sec x = 1 \)
• D. \( \sec x + \cosec x = 2 \)

Hint:

Always check the domain of the trigonometric functions in the options. Solutions like \( \sin x = 0 \) make \( \cot x \) and \( \cosec x \) undefined immediately eliminating those options.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:

We first solve the irrational trigonometric equation for \( x \), then substitute the solution into the given options to check for validity.
Step 2: Key Formula or Approach:

1. Isolate the square root and square both sides (ensure LHS \( \ge 0 \)). 2. Use \( \sin^2 x = 1 - \cos^2 x \).
Step 3: Detailed Explanation:

Equation: \( \sqrt{6 - 5\cos x + 7\sin^2 x} = \cos x \). For solution to exist, \( \cos x \ge 0 \). Squaring both sides: \[ 6 - 5\cos x + 7(1-\cos^2 x) = \cos^2 x \] \[ 6 - 5\cos x + 7 - 7\cos^2 x = \cos^2 x \] \[ 13 - 5\cos x = 8\cos^2 x \] \[ 8\cos^2 x + 5\cos x - 13 = 0 \] Factorizing or using quadratic formula for \( \cos x \): \( 8\cos^2 x + 13\cos x - 8\cos x - 13 = 0 \) \( \cos x(8\cos x + 13) - 1(8\cos x + 13) = 0 \) \( (\cos x - 1)(8\cos x + 13) = 0 \) Since \( |\cos x| \le 1 \), \( \cos x = -13/8 \) is rejected. Thus, \( \cos x = 1 \). If \( \cos x = 1 \), then \( \sin x = 0 \), \( \tan x = 0 \), \( \sec x = 1 \). Note: \( \cot x \) and \( \cosec x \) are undefined. Checking Options: (A) \( \tan x + \cot x \): Undefined. (B) \( \cot x + \cosec x \): Undefined. (C) \( \tan x + \sec x = 0 + 1 = 1 \). (Valid) (D) \( \sec x + \cosec x \): Undefined.
Step 4: Final Answer:

The equation satisfies \( \tan x + \sec x = 1 \).