Trig. Eqt.

Graphically I can see that a can be any real number.
-∞ < a < ∞

2a tan x + a - 18 sin x = 0

2a tan x + a = 18 sin x

a(2 tan x + 1) = 18 sin x

a = (18 sin x)/(2 tan x + 1)

a = (18 sin x cos x)/(2 sin x + cos x)

Try to show that this expression can attain any real value.

2015-07-29 03:26:48 補充：
2a tan x + a - 18 sin x = 0

2a tan x + a = 18 sin x

a(2 tan x + 1) = 18 sin x

a = (18 sin x)/(2 tan x + 1)

Note that a is undefined when tan x = -1/2
which occurs when for example x = π - tan⁻¹(1/2) ≈ 2.67795.

Consider when tan x is close to -1/2 (take the case when x is close to π - tan⁻¹(1/2) which is in Quadrant II, so sin x is positive).

When tan x tends to (-1/2)- (i.e., x tends to 2.67795-),
(2 tan x + 1) tends to 0-, so a tends to -∞.
When tan x tends to (-1/2)+ (i.e., x tends to 2.67795+),
(2 tan x + 1) tends to 0+, so a tends to +∞.

Note also the fact that (the local maximum of a) > (the local minimum of a).
(Read the next post you asked for the local maximum, in the Comments.)

Therefore, a can take any real number.

As a result, for x to be real, the range of a is (-∞, +∞).
The range of positive value of a for x to be real (as requested) is (0, +∞).


2015-07-29 03:27:32 補充：
Please also associate this post with the one below:

https://hk.knowledge.yahoo.com/question/question?qid=7015072700006

(for the local maximum of a, please read the Comments there.)