Thread: Equation

2sin3X = 2^1/2

for 0 ≤ x < 360

I narrowed down to 3X = 45 --> x = 15 and 3X = 135 --> x = 45; but my instructor says I am still missing 4.

Originally Posted by windia

2sin3X = 2^1/2

for 0 ≤ x < 360

I narrowed down to 3X = 45 --> x = 15 and 3X = 135 --> x = 45; but my instructor says I am still missing 4.

sine is periodic. Therefore (working in degrees) 3x = 45 + 360n or 3x = 135 + 360n where n is an integer. Solve for x and then use all values of n that give solutions within the given domain.