How Do You Solve Sinx=1/2?

     

So my main question is: Should the general solution be $npi$ + $(-1)^n$$frac7pi6$ or $npi$ + $(-1)^n$$frac11pi6$?

Apparently, WolframAlpha uses $frac7pi6$ for the general solution. My apologies lớn you if you found this question a bit silly. I appreciate your time and effort.

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There is no difference, since$$3pi+(-1)^3frac7pi6=frac11pi6quad extandquad3pi+(-1)^3frac11pi6=frac7pi6.$$So, you get the same solutions in both cases.


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See if this helps:

We know $sin heta$ is negative in quadrants 3 & 4:

Therefore, for Quadrants 3 & 4, these formulas can be used respectively:$$ extQuadrant 3: (2n + 1)pi + heta$$$$ extQuadrant 4: 2pi(n + 1) - heta$$where $n$ denotes, the rotation (repeat; as the cycle repeats ever $2pi$) count và $ heta$ is, well, your angle (which, the base is, $fracpi6$)


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