solve the equation cos x + sin x = \(\frac{1}{cos x - sinx}\) for values...

Mathematics

JAMB 1998

solve the equation cos x + sin x = \(\frac{1}{cos x - sinx}\) for values of such that 0 \(\leq\) x < 2\(\pi\)

A. \(\frac{\pi}{2}\), \(\frac{3\pi}{2}\)
B. \(\frac{\pi}{3}\), \(\frac{2\pi}{3}\)
C. 0, \(\frac{\pi}{3}\)
D. 0, \(\pi\)

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Correct Answer: Option D

Explanation

cos x + sin x = \(\frac{1}{cos x - sinx}\)

= (cosx + sinx)(cosx - sinx) = 1

= cos\(^2\)x - sin\(^2\)x = 1

cos\(^2\)x - (1 - cos\(^2\)x) = 1

= 2cos\(^2\)x = 2

cos\(^2\)x = 1

= cosx = \(\pm\)1

x = cos\(^{-1}\) (\(\pm\)1)

= 0, \(\pi\) \(\frac{3}{2}\pi\), 2\(\pi\)

(possible solution)