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, \(\pi\)
Verified Correct Answer: D.
Step-by-Step Mathematical Solution
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)
Source: Myschool Past Questions Archive (Record #30729)