How do you solve 2 sin x + 1 = 0?

1 Answer
Truong-Son N.
May 2, 2015

Subtract 1, divide by 2.

sinx = -1/2

sinx = pm1/2 at 30^o, 150^o, 210^o, and 330^o.
sinx = -1/2 at 210^o and 330^o.

So, you have two answers. Really though, you have infinite answers if you consider that there are equivalent angles at every pm360^o. What you get, then, is:

210^o/180^o*pi = (7pi)/6

330^o/180^o*pi = (11pi)/6

Therefore, sinx = -1/2 at (7pi)/6, (11pi)/6, (19pi)/6, (23pi)/6, etc.

Notice how (11pi)/6 = 2pi - pi/6, and (7pi)/6 = 0 + (7npi)/6. Recall how sin 0 = sin 2pi = sin 2npi, where nin ZZ (n is in the set of integers). Thus, you can write:

x = 2npi-pi/6

x = 2npi+(7pi)/6

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