Question

TÌM TXĐ:

a. y=\(\dfrac{cos2x}{1-sin2x}\)

b. y= cosx- tan(4x+\(\dfrac{\pi}{3}\)) +5

c. y=cos\(\dfrac{x}{3}\)- \(\dfrac{3}{1+sin2x}\)+ \(\dfrac{2}{3}\)

Answer 1

a/ Điều kiện: 1 - sin2x \(\ne\) 0
<=> sin2x \(\ne1\)
<=> \(x\ne\dfrac{\pi}{4}+k\dfrac{\pi}{2}\)
TXĐ: D = R\ {\(\dfrac{\pi}{4}+k\dfrac{\pi}{2}\)}

Answer 2

b. ĐKXĐ cos(4x+\(\dfrac{\pi}{3}\)) \(\ne\)0 => 4x+\(\dfrac{\pi}{3}\)= \(\dfrac{\pi}{2}\)+k\(\pi\) => x=\(\dfrac{\pi}{24}\)+k\(\dfrac{\pi}{4}\),k\(\in\)Z

==> TXĐ: D= R\ { \(\dfrac{\pi}{24}\)+k\(\dfrac{\pi}{4}\),k\(\in\)Z }

Answer 3

c. ĐKXĐ : 1+ sin2x \(\ne\) 0 => sin2x \(\ne\) -1 => 2x= -\(\dfrac{\pi}{2}\)+k2\(\pi\)

=> x= -\(\dfrac{\pi}{4}\)+k\(\pi\)

TXĐ : D = R \ { -\(\dfrac{\pi}{4}\)+k\(\pi\) }