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 \( \tan x + \sec x = 1 \).