Find the definition domain of function y = cosx

Find the definition domain of function y = cosx

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Find the definition domain of function y = LG (SiNx cosx)

According to the meaning of the question, SiNx cosx > 0, SiNx > cosx is divided into cases. When cosx = 0, SiNx = 1, x = π / 2 + 2K π, when cosx > 0 (x is in quadrant 1,4), SiNx > cosx is TaNx > 1 (x is in quadrant 1,3). At this time, X can only be in the first quadrant. It can be known from the image that π / 4 + 2K π < x < π / 2 + 2K π when cosx < 0 (x is in quadrant 2,3

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SiNx > cosx sketch
π/4+2kπ

Function y = lgsinx+ cosx−1 The domain of 2 is __

(1) To make a function meaningful, you must have
sinx＞0
cosx−1
2≥0 ，
Namely
sinx＞0
cosx≥1
two ，
Solution
2kπ＜x＜π+2kπ
−+2kπ≤x≤π
3+2kπ （k∈Z），
∴2kπ＜x≤π
3+2kπ，k∈Z，
The definition domain of the function is {x|2k π < x ≤ π
3+2kπ，k∈Z}．
So the answer is: {x|2k π < x ≤ π
3+2kπ，k∈Z}

The function f (x) = LG (SiNx cosx) is defined as () A. {x|2kπ−3π 4＜x＜2kπ+π 4，k∈Z} B. {x|2kπ+π 4＜x＜2kπ+5π 4，k∈Z} C. {x|kπ−π 4＜x＜kπ+π 4，k∈Z} D. {x|kπ+π 4＜x＜kπ+3π 4，k∈Z}

From SiNx cosx > 0, we get
sinx＞cosx．
The solution is 2K π + π
4＜x＜2kπ+5π
4，k∈Z．
Therefore, the definition domain of the original function is {x|2k π + π
4＜x＜2kπ+5π
4，k∈Z}．
Therefore, B

Domain of function y = LG (SiNx cosx / SiNx + cosx) help! > oo

(sinx-cosx)/(sinx+cosx)>0
Cosx = 0, obviously the solution, x = k π + π / 2
cosx0,(tgx-1)/(tgx+1)>0---> tgx>1 or tgx kπ+π/4