solve_trig_factor_use.py

# examples how we can use factor to reduce number of imageset :

# current master branch
In [110]: solveset(4*sin(x)**3  + 2*sin(x)**2 - 2*sin(x) - 1)  # ------------ eq(1)
Out[110]: 
⎧        7⋅π        ⎫   ⎧        11⋅π        ⎫   ⎧        5⋅π        ⎫   ⎧        3⋅π        ⎫   ⎧        7⋅π        ⎫   ⎧        π        ⎫
⎨2⋅n⋅π + ─── | n ∊ ℤ⎬ ∪ ⎨2⋅n⋅π + ──── | n ∊ ℤ⎬ ∪ ⎨2⋅n⋅π + ─── | n ∊ ℤ⎬ ∪ ⎨2⋅n⋅π + ─── | n ∊ ℤ⎬ ∪ ⎨2⋅n⋅π + ─── | n ∊ ℤ⎬ ∪ ⎨2⋅n⋅π + ─ | n ∊ ℤ⎬
⎩         6         ⎭   ⎩         6          ⎭   ⎩         4         ⎭   ⎩         4         ⎭   ⎩         4         ⎭   ⎩        4        ⎭

In [123]: solveset((4*sin(x)**3  + 2*sin(x)**2 - 2*sin(x) - 1).rewrite(exp))
Out[123]: 
⎧        5⋅π        ⎫   ⎧        π        ⎫   ⎧        3⋅π        ⎫   ⎧        π        ⎫   ⎧        3⋅π        ⎫   ⎧        π        ⎫
⎨2⋅n⋅π - ─── | n ∊ ℤ⎬ ∪ ⎨2⋅n⋅π - ─ | n ∊ ℤ⎬ ∪ ⎨2⋅n⋅π + ─── | n ∊ ℤ⎬ ∪ ⎨2⋅n⋅π + ─ | n ∊ ℤ⎬ ∪ ⎨2⋅n⋅π - ─── | n ∊ ℤ⎬ ∪ ⎨2⋅n⋅π - ─ | n ∊ ℤ⎬
⎩         6         ⎭   ⎩        6        ⎭   ⎩         4         ⎭   ⎩        4        ⎭   ⎩         4         ⎭   ⎩        4        ⎭

# first factor the trig eq
In [112]: fact = factor_list(4*sin(x)**3  + 2*sin(x)**2 - 2*sin(x) - 1)  # ------- eq(1) factors

In [113]: fact
Out[113]:
⎛   ⎡                   ⎛     2          ⎞⎤⎞
⎝1, ⎣(2⋅sin(x) + 1, 1), ⎝2⋅sin (x) - 1, 1⎠⎦⎠

# see the factors exp form(factor them also to remove extra exp)
In [114]: factor((2*sin(x) + 1).rewrite(exp))   # -------eq(1), factor (1)
Out[114]:
 ⎛   2⋅ⅈ⋅x    ⅈ⋅x    ⎞  -ⅈ⋅x
-⎝ⅈ⋅ℯ      - ℯ    - ⅈ⎠⋅ℯ

In [115]: factor((2*sin(x)**2 - 1).rewrite(exp))  # -------eq(1), factor (2)
Out[115]:
 ⎛ 4⋅ⅈ⋅x    ⎞  -2⋅ⅈ⋅x
-⎝ℯ      + 1⎠⋅ℯ
──────────────────────
          2

# both outer exp will not give any solution
In [116]: solveset(exp(-I*x)) #  --------- outside bracket - factor (1)
Out[116]: ∅

In [117]: solveset(exp(-2*I*x))  # --------- outside bracket - factor (2)
Out[117]: ∅

# these both will  contribute in soln
In [118]: solveset(I*exp(2*I*x) - exp(I*x) - I)  # --------- inside bracket - factor (1)
Out[118]:
⎧        5⋅π        ⎫   ⎧        π        ⎫
⎨2⋅n⋅π - ─── | n ∊ ℤ⎬ ∪ ⎨2⋅n⋅π - ─ | n ∊ ℤ⎬
⎩         6         ⎭   ⎩        6        ⎭


In [122]: solveset(exp(4*I*x) + 1)  # --------- inside bracket - factor (2)
Out[122]:
⎧n⋅π   π        ⎫
⎨─── + ─ | n ∊ ℤ⎬
⎩ 2    4        ⎭

# now compare if we direclty solve the trig factor

# this looks same as above
In [120]: solveset(2*sin(x) + 1)  # ---------current branch solves without factoring, so soln is similar - factor (1)
Out[120]:
⎧        7⋅π        ⎫   ⎧        11⋅π        ⎫
⎨2⋅n⋅π + ─── | n ∊ ℤ⎬ ∪ ⎨2⋅n⋅π + ──── | n ∊ ℤ⎬
⎩         6         ⎭   ⎩         6          ⎭

# but this contains many imageset that is reduced above
In [121]: solveset(2*sin(x)**2  - 1) # ---------current branch solves without factoring, so soln is larger than Out[122] - factor (2)
Out[121]:
⎧        5⋅π        ⎫   ⎧        3⋅π        ⎫   ⎧        7⋅π        ⎫   ⎧        π        ⎫
⎨2⋅n⋅π + ─── | n ∊ ℤ⎬ ∪ ⎨2⋅n⋅π + ─── | n ∊ ℤ⎬ ∪ ⎨2⋅n⋅π + ─── | n ∊ ℤ⎬ ∪ ⎨2⋅n⋅π + ─ | n ∊ ℤ⎬
⎩         4         ⎭   ⎩         4         ⎭   ⎩         4         ⎭   ⎩        4        ⎭

# so overall we have (if we use factor correctly, less imagesets we get)
# so ans from factor(1) and factor(2) is here(which is simpler than current master) : 
⎧        5⋅π        ⎫   ⎧        π        ⎫
⎨2⋅n⋅π - ─── | n ∊ ℤ⎬ ∪ ⎨2⋅n⋅π - ─ | n ∊ ℤ⎬ ∪
⎩         6         ⎭   ⎩        6        ⎭

⎧n⋅π   π        ⎫
⎨─── + ─ | n ∊ ℤ⎬
⎩ 2    4        ⎭

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