Solve sin(θ) = x in the range a ≤ θ ≤ b

Question

Challenge

Given a number, x where -1 ≤ x ≤ 1, and the integers a and b, where b > a, solve the equation sin(θ) = x for θ, giving all solutions in the range a° ≤ θ ≤ b°.

Rules

x will have a maximum of three decimal places of precision. The output, θ, must also be to three decimal places, no more, no less.

Your output may be as a list or separated output.

You may use built-in​ trig functions.

You may not use radians for a and b.

You will never be given invalid input.

Output may be in any order.

Examples

x, a, b => outputs
0.500, 0, 360 => 30.000, 150.000
-0.200, 56, 243 => 191.537
0.000, -1080, 1080 => 0.000, 180.000, 360.000, 540.000, 720.000, 900.000, 1080.000, -1080.000 -900.000, -720.000, -540.000, -360.000, -180.000

Winning

Shortest code in bytes wins.

Answer 1

Fortran 95 (gfortran), 180 bytes

#define G >=i)write(*,'(f9.3)')
#define H read(*,*)
program a
H x
H i
H j
o=ASIN(x)*57.2957
p=o-180+o
q=o-360*999
do
r=q-p
if(r<=j.and.r G r
if(q>=j)exit
if(q G q
q=q+360
enddo
end

Structure ungolfed:

program a
        implicit none
        real :: x
        integer :: i
        integer :: j
        real :: o
        real :: r
        real :: q
        real :: p

        read(*,*) x
        read(*,*) i
        read(*,*) j

        o=ASIN(x)*57.2957
        p=o-180+o
        q=o-360*999

        do
                r=q-p
                if(r<=j.and.r >=i) then
                        write(*,'(f9.3)') r
                endif
                if(q>=j) then
                        exit
                endif
                if(q >=i) then
                        write(*,'(f9.3)') q
                endif
                q = q + 360
        end do
end program a

Answer 2

Mathematica, 60 bytes

NumberForm[t/.NSolve[Sin[t/180Pi]==#&&#2<=t<=#3,{t}],{9,3}]&

input

[0, -1080, 1080]

output

{-1080.000,-900.000,-720.000,-540.000,-360.000,-180.000,0.000,180.000,360.000,540.000,720.000,900.000,1080.000}

Answer 3

APL, 51 46 40 39 bytes

_{3 bytes saved thanks to @KritixiLithos}

_{6 7 bytes saved thanks to @Adám}

{3⍕⍵/⍨1E¯6>|⎕-1○○⍵÷180}⊢+1E3÷⍨(⍳1001×-)

Called as a dyad with a as left argument and b as right argument, prompts for x.

Requires ⎕IO←0.

How?

                         ⊢+1E3÷⍨(⍳1001×-)  ⍝ build the range a to b with step of 0.001
                                  1001×-   ⍝ 1001 * (b - a)
                                 ⍳         ⍝ range
                           1E3÷⍨           ⍝ divide every element by 1000
                         ⊢+                ⍝ add a back

{3⍕⍵/⍨1E¯6>|⎕-1○○⍵÷180}                  ⍝ filter the solutions
                  ○⍵÷180                   ⍝ convert to radian - π * ⍵ / 180
                1○                          ⍝ compute sine
             |⎕-                           ⍝ distance from x
       1E¯6>                                ⍝ small enough
    ⍵/⍨                                    ⍝ compress with the original list
 3⍕                                        ⍝ format to 3 decimal places

Answer 4

PHP>=7.1, 88 Bytes

for([,$s,$f,$t]=$argv;$f<=$t;$f+=.001)round(sin(deg2rad($f)),5)!=$s?:printf("%.3f_",$f);

Online Version

Answer 5

Python 2, 100 bytes

from math import*
x,a,b=input()
while a<=b:
    if round(sin(pi*a/180),5)==x:print a
    a=round(a+.001,3)

Try it online

Answer 6

Axiom, 266 215 210 bytes

g(y,a,b)==(r:List Float:=[];t:=y+truncate((a-y)/360)*360;repeat(t>b=>break;if t>=a then r:=append(r,[t]);t:=t+360);r)
f(x,a,b)==(abs(x)>1 or a>b=>[];y:=asin(x)*180/%pi;z:=180-y;sort(append(g(y,a,b),g(z,a,b))))

ungolf and test

-- g/180=r/pi  => r=pi*g/180
--y+k*360=a
--    a-y
-- k=------
--    360
-- find all angles of the same class of angles [y+k*360] in interval [a,b]
gg(y,a,b)==
   r:List Float:=[]
   t:=y+truncate((a-y)/360)*360 --truncate(1.9)=1, truncate(-3.1)=-4 is ok
   repeat
     t>b=>break
     if t>=a then r:=append(r,[t])
     t:=t+360
   r

-- ff returns the list of solution y of sin(y)=x with y in the interval [a,b]
ff(x,a,b)==
   abs(x)>1 or a>b =>[]
   y:=asin(x)*180/%pi -- z and y are the only 2 solutions in one 360 Len interval 
   z:=180-y
   sort(append(gg(y,a,b),gg(z,a,b)))


(6) -> f(0.5, 0, 360)
   (6)  [30.0,150.0]
                                                         Type: List Float
(7) -> f(-0.2, 56, 243)
   (7)  [191.5369590328 1548769]
                                                         Type: List Float
(8) -> f(0.0, -1080, 1080)
   (8)
   [- 1080.0, - 900.0, - 720.0, - 540.0, - 360.0, - 180.0, 0.0, 180.0, 360.0,
    540.0, 720.0, 900.0, 1080.0]

(14) -> m:=f(-0.1, -2035, -243)
   (14)
   [- 1974.2608295227 33214, - 1805.7391704772 66786, - 1614.2608295227 33214,
    - 1445.7391704772 66786, - 1254.2608295227 33214, - 1085.7391704772 66786,
    - 894.2608295227 332137, - 725.7391704772 667863, - 534.2608295227 332137,
    - 365.7391704772 667863]
                                                         Type: List Float
(15) -> map(x+->sin(%pi*x/180), m)
   (15)
   [- 0.0999999999 9999999987, - 0.1000000000 0000000016,
    - 0.0999999999 9999999986 2, - 0.1000000000 0000000028,
    - 0.0999999999 9999999985 4, - 0.1000000000 0000000018,
    - 0.0999999999 999999999, - 0.1000000000 0000000019,
    - 0.0999999999 9999999989 2, - 0.1000000000 0000000014]
                                                         Type: List Float

Answer 7

R, 74 bytes

function(x,a,b,u=seq(a,b,.001))sprintf("%.3f",u[abs(sinpi(u/180)-x)<1e-6])

Try it online!

The tolerance 1e-6 was chosen to fit the second test case, so is somewhat arbitrary. This answer takes advantage of R's vectorized functions.