Calculus I - Trigonometry problem

[Taibhrigh]

For a homework assignment, I was given the problem:

Solve for u in the interval [0,2π) 2sin3u + √(2) = 0

I tried, and my work was:

2sin(2u + u) + √(2) = 0
2(sin2ucosu + cos2usinu) + √(2) = 0
2sinucos²u + 2cos2usinu + √(2) = 0
sinucos²u + cos2usinu = √(2)/2

That's as far as I got before completely confusing myself. (I have, obviously, not taken Trig before this Calculus course.)
To get credit, though, I have to figure out how to solve this, and the lady at the resource center doesn't remember any Calculus. Can anyone help me with solving this, please?

 

[soroban]

Hello, Taibhrigh!

You're making hard work for yourself . . .

Solve for \(\displaystyle u\) in the interval \(\displaystyle [0,\,2\pi):\:2\sin(3u)\,+\,\sqrt{2}\:=\:0\)

We have: \(\displaystyle \:\sin(3u)\:=\:\L-\frac{\sqrt{2}}{2}\)

Then: \(\displaystyle \L\:3u\:=\:\frac{5\pi}{4},\:\frac{7\pi}{4},\:\frac{13\pi}{4},\:\frac{15\pi}{4},\:\frac{21\pi}{4},\:\frac{23\pi}{4}\)

Therefore: \(\displaystyle \L\:u \;=\;\frac{5\pi}{12},\:\frac{7\pi}{12},\:\frac{13\pi}{12},\:\frac{5\pi}{4},\frac{7\pi}{4},\:\frac{23\pi}{12}\)