asinx+bcosx=i*sin(x+q)

Where tanq=b/a(q is the value of angle and radian), and I = (a 2+b 2) is under the root sign.

As can be seen from the figure, the function y is 60 degrees to the right of sinx +m*360 degrees (the minimum positive period of sinx is 360 degrees, and m is an integer).

So q=-60 degrees +m*360 degrees (function translation law: left plus right minus, m is an integer).

The formula of b/a =-k = tanq =- radical number 3 ** 1 is derived.

Let the maximum value of the function in the diagram be I (you should be able to see it, otherwise this problem can't be done)

So i= (A 2+B 2) * * 2 under the radical sign.

** 1 formula **2 formula is solved simultaneously to get a and b.

(b)2kcosx-2sinx=- 1

Auxiliary angle formula

sin(x+n)=- 1

tan n= - 1/k

(if K > 0)K obtained in (a)) k takes four quadrant angles.

(if k

X+n=-90 degrees +360*m (m is an integer)