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Answer:
[tex] \displaystyle \large{ x = \frac{7\pi}{6} + 2\pi k , \frac{11\pi}{6} + 2\pi k \: \: \: \: \: (k \in \Z)}[/tex]
Step-by-step explanation:
First, isolate sinx.
[tex] \displaystyle \large{2 \sin x + 1 = 0}[/tex]
Subtract both sides by 1; after that, divide both sides by 2.
[tex] \displaystyle \large{2 \sin x + 1 - 1 = 0 - 1} \\ \displaystyle \large{2 \sin x = - 1} \\ \displaystyle \large{ \frac{2 \sin x}{2} = \frac{ - 1}{2} } \\ \displaystyle \large{ \sin x = - \frac{1}{2} }[/tex]
Now recall a unit circle. Sine is only negative in Q3 and Q4.
1/2 is 30° or π/6 for sine.
Finding the Q3 and Q4 angles.
Q3
Using π + π/6 = 6π/6 + π/6 = 7π/6
Q4
Using 2π - π/6 = 12π/6 - π/6 = 11π/6
Now we know that in Q3, sinx = -1/2 is 7π/6 and in Q4, it's 11π/6.
Thus:-
[tex] \displaystyle \large{ x = \frac{7\pi}{6} , \frac{11\pi}{6} }[/tex]
But since you did not specific the interval, assume that we are finding general solutions.
From Identity:
[tex] \displaystyle \large{ \sin x = x + 2\pi k \: \: \: (k \in \Z)}[/tex]
Therefore since x = 7π/6 and x = 11π/6:
[tex] \displaystyle \large{ x = \frac{7\pi}{6} + 2\pi k , \frac{11\pi}{6} + 2\pi k}[/tex]
where k is any integers.