Maths about graph

Q1. When x = 0, y = 2, so 2 = a sin 0 + b cos 0 - 1, 2 = b - 1, b = 3.
When x = 90, y = 3, so 3 = a sin 90 + 3 cos 90 - 1 = a - 1, so a = 4.
That is y = 4 sin x + 3 cos x - 1.
For equation 4 sin x + 3 cos x + 2 = 0
4 sin x + 3 cos x - 1 + 3 = 0
4 sin x + 3 cos x - 1 = - 3,
that means the intersecting points of y = 4 sin x + 3 cos x - 1 and y = - 3 will be the solution to the equation.




2009-08-13 10:48:42 補充：
For 4 sin x + 3 cos x < - 3, so 4 sin x + 3 cos x - 1 < - 4, that is y < - 4, that means the section of the curve that is below y = - 4 will be the required range of x.

2009-08-13 13:41:16 補充：
Q25. Adding the 2 equations together, we get sin x = (a + b)/2. Subtracting the 2 equations we get cos x = (a - b)/2. sin^2 x + cos^2 x = 1 = [(a + b)/2]^2 + [(a - b)/2]^2 = 2(a^2/4 + b^2/4) = (a^2 + b^2)/2.

2009-08-13 13:47:09 補充：
Q25. cont'd. If a = (sqrt5)/2, so [5/4 + b^2]/2 = 1, b^2 = 2 - 5/4 = 3/4, so b =
(sqrt3)/2 [ - sqrt3/2 is rej because b > 0]. So sin x = ( sqrt 5 / 2 + sqrt 3 / 2)/2
= (sqrt 5 + sqrt 3)/4 = 0.9920296, so x = arcsin 0.9920296 = 82.8 degree.

2009-08-13 13:55:53 補充：
Q26. (i) x = arcsin (2/5) = 23.58 degree and (180 - 23.58) = 156.42 degree.
(ii) x = arccos (4/7) = 55.15 degree and (360 - 55.15) = 304.85 degree.
(b) Let sin(x/3) = x and cos(x - 20) = y, so f(x) = 8 - 20x - 14y + 35xy
= 4(2 - 5x) - 7y(2 - 5x) = (4 - 7y)(2 - 5x) = [4 - 7cos(x - 20)][2 - 5 sin(x/3)].

2009-08-13 13:59:55 補充：
Q26 cont'd. (a) 4 - 7 cos(x - 20) = 0, x - 20 = arccos(4/7) = 55.15 and 304.85, so x = 75.15 and 324.85. (b) 2 - 5 sin(x/3) = 0, x/3 = arcsin(2/5) = 23.58 and 156.42, so x = 70.74 and 469.26 = 469.26 - 360 = 109.26.

2009-08-13 14:03:15 補充：
Sorry, 469.26 should be rejected, so 109.26 is wrong, please delete this one.

Thx~but how about Q.25-26?