I recently took the trigonometric function and radian system in compulsory four of senior one mathematics, and I feel a little confused in some places, that is, when both sets are written in radian system. First question: According to your method, the third quadrant can be changed to (2K+ 1)π+π/4 instead of (2K+ 1)π-π/4 and then merged into Kπ+π/4. This kind of problem must first find the right period. The period of the diagonal of the first and third quadrant angles is obviously π, not 2π. So you got the wrong cycle, which made the topic very troublesome. Find the period π and add the simplest value π/4. The set expression is Kπ+π/4.

The second question: First of all, A is a periodic region, starting from a simpler B. B = x 2-2x-3 <; =0, that is, the value of X is between two solutions of the equation X 2-2x-3 = 0, that is,-1

Then look at the range of k values of a that may intersect with B. This is easy to get only when k=0 has an intersection in region A. namely

-π/3 & lt； = x < = 2π/3, the last comparison range. Because -π/3 is about -3. 14/3, it is less than-1; 2π/3 is approximately equal to 2/3*3. 14 and less than 3. If the intersection part is taken, the intersection of A and B is-1