solve the equation cos x + sin x = \(\frac{1}{cos x – sinx}\) for values of such that 0 \(\leq\) x < 2\(\pi\)
- \(\frac{\pi}{2}\), \(\frac{3\pi}{2}\)
- \(\frac{\pi}{3}\), \(\frac{2\pi}{3}\)
- 0, \(\frac{\pi}{3}\)
-
0, \(\pi\)
The correct answer is: D
Explanation
cos x + sin x = \(\frac{1}{cos x - sinx}\)
= (cosx + sinx)(cosx - sinx) = 1
= cos2x + sin2x = 1
cos2x - (1 - cos2x) = 1
= 2cos2x = 2
cos2x = 1
= cosx = \(\pm\)1 = x
= cos-1x (\(\pm\), 1)
= 0, \(\pi\) \(\frac{3}{2}\pi\), 2\(\pi\)
(possible solution)