The number of intersections of the images of functions y = SiNx and y = TaNx on [- 2 π, 2 π] is () A. 3 B. 5 C. 7 D. 9

The number of intersections of the images of functions y = SiNx and y = TaNx on [- 2 π, 2 π] is () A. 3 B. 5 C. 7 D. 9

Method 1: image method, draw y = SiNx and y = TaNx in the same coordinate system
The image on [0,2 π] shows that the image of function y = SiNx and y = TaNx has five intersections on [- 2 π, 2 π],
Therefore, B
Method 2: solve the equation SiNx = TaNx, that is, TaNx (cosx-1) = 0,
‡ TaNx = 0 or cosx = 1, ∵ x ∈ [- 2 π, 2 π],
‡ x = 0, ± π, ± 2 π, so there are five solutions,
Therefore, B

Let me ask you a few more questions. Ha, how many intersection points do the function TaNx and the function y = SiNx have in the interval (- 270 °, 270 °), The answer is three, but I draw five intersections. I don't know what went wrong. I'm a little incomplete in TaNx chapter

There are only three points, all of which are 0 points, i.e. x = - 180 °, x = 0, x = 180 °;
Key knowledge: in the (0,90 °) interval, SiNx TaNx = SiNx SiNx / cosx = (sinxcosx SiNx) / cosx = SiNx (sinx-1) / cosx
Obviously, because 0

Function y = SiNx + TaNx absolute value SiNx TaNx absolute value. X belongs to (pie ÷ 2,3 pie divided by 2). What is the value range? emergency

Y = SiNx + TaNx - |sinx tanx| when x is (π / 2, π), SiNx > TaNx
Y = SiNx + TaNx SiNx + TaNx = 2tanx y range (negative infinity, 0)
SiNx when x is (π, 3 π / 2)

Function y = |cosx| cosx+tanx |The value range of TaNx | is __

When the angle is the angle in the first quadrant, y = 1 + 1 = 2,
When the angle is the angle of the second quadrant, y = - 1-1 = - 2,
When the angle is the angle of the third quadrant, y = - 1 + 1 = 0,
When the angle is the angle of the fourth quadrant, y = 1-1 = 0,
It can be seen that the value range of the function is {- 2, 0, 2},
So the answer is: {- 2, 0, 2}

What is the value range of the function y = | cosx| / cosx + | tanx| / TaNx?

In the first and fourth quadrants, the cosine is positive, and the tangent of the first and third quadrants is positive (x is not an odd multiple of π / 2)
So in the first quadrant y = 2, the second quadrant y = - 2, and the third and fourth quadrants y = 0
Therefore, the value range is {- 2,0,2}

The parity of the function y = x ^ 2 + 1 / (cosx-2) is

Even function, y (- x) = (- x) ^ 2 + 1 / (COS (- x) - 2), cos (- x) = cosx, so y (- x) = y (x). So it is even