SOLVE THIS EQUATION HELP

sin2x=tanx
looks simple, yet confusing
interval [0, 360]degrees

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                      sin(2x) = tan(x)            ← Convert to sine & cosine
             2sin(x)cos(x) = sin(x)/cos(x)            ← used identity sin2θ = 2sinθcosθ
           2sin(x)cos²(x) = sin(x)            ← multiplied both sides by cos(x)
2sin(x)cos²(x) - sin(x) = 0            ← Note: I didn't divide both sides by sine because
                                                                  that might loose part of the solution
     sin(x)[2cos²(x) - 1] = 0

Possible solutions are for:
➊ sin(x) = 0
           x = 0°, 180°, 360° ✔            ← All three work in the original equation

➋ 2cos²(x) - 1 = 0
           cos²(x) = ½
           cos(x) = ±√(½)
           cos(x) = ±(√2)/2            ← Note: cos(45°) = (√2)/2
                    x = 45°, 135°, 225°, 315° ✔            ← All three work in the
                                                                                  original equation

                        ANSWER
   x = 0°, 45°, 180°, 135°, 225°, 315°, 360°

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double-angle formula
sin(2x) = 2sin(x)cos(x)

2sin(x)cos(x) = tan(x)
2sin(x)cos(x) = sin(x)/cos(x)
2cos(x) = 1/cos(x)
2cos²(x) = 1
2cos²(x) = cos²(x) + sin²(x) pythagorean identity
cos²(x) = sin²(x)
x = 45°, 135°, 225°, 315°

sin 90 = 1
tan 45 = 1
x = 45 degrees

sin 270 = -1
tan 135 = -1
x = 135 degrees
[45, 135]