+1904 \(2{sin}^{2}(x)+3sin(x) = -1\)
Let \(sin(x)=x\)
\(2{x}^{2}+3x=-1\)
\({2x}^{2}+3x+1=0\)
\((2x+1)(x+1)=0\)
\((2sin(x)+1)(sin(x)+1)=0\)
\(2sin(x)+1=0\) \(sin(x)+1=0\)
\(2sin(x)+1=0\)
\(2sin(x)=-1\)
\(sin(x)=\frac{-1}{2}\)
\(sin(x)=-\frac{1}{2}\)
\({sin}^{-1}(sin(x))={sin}^{-1}(-\frac{1}{2})\)
\(x=-\frac{\pi}{6}\)
Because \(-\frac{\pi}{6}\) is not the only answer to the first half of the equation, if you add \(2\pi\) over and over again forever, you will get infinite answers. Represent this as \(x=-\frac{\pi}{6}+2\pi n\). \(n=\)numer of times around the unit circle (positive integer equals counterclockwise the around unit circle and negetive integer equals clockwise around the unit circle).
Because \({sin}^{-1}\)only gives one answer to the first half of the equation, to find the other answer, look at the unit circle and see where \(sin\) also equals \(-\frac{1}{2}\) and when you do that, you get that \(x=-\frac{5\pi}{6}.\)
Because \(-\frac{5\pi}{6}\) is not the only answer to the first half of the equation, if you add \(2\pi\) over and over again forever, you will get infinite answers. Represent this as \(x=-\frac{5\pi}{6}+2\pi n\). \(n=\)numer of times around the unit circle (positive integer equals counterclockwise the around unit circle and negetive integer equals clockwise around the unit circle).
If having a negetive fraction at the begining of each answer looks weird, you can add \(2\pi\) to the negetive fraction to change the negetiive fraction to a positive fraction. When you do that to the two answers to the first part of the equation, you get \(x=\frac{11\pi}{6}+2\pi n\) and \(x=\frac{7\pi}{6}+2\pi n\)
The two answers for the first half of the equation is \(x=-\frac{\pi}{6}+2\pi n\) \((x=\frac{11\pi}{6}+2\pi n)\) and \(x=-\frac{5\pi}{6}+2\pi n\) \((x=\frac{7\pi}{6}+2\pi n).\)
\(sin(x)+1=0\)
\(sin(x)=-1\)
\({sin}^{-1}(sin(x))={sin}^{-1}(-1)\)
\(x=-\frac{\pi}{2}\)
Because \(-\frac{\pi}{2}\) is not the only answer to the second half of the equation, if you add \(2\pi\) over and over again forever, you will get infinite answers. Represent this as \(x=-\frac{\pi}{2}+2\pi n\). \(n=\)numer of times around the unit circle (positive integer equals counterclockwise the around unit circle and negetive integer equals clockwise around the unit circle).
If you look at the unit circle, you will see that there is only one answer to the second half of the equation. The only answer is \(x=-\frac{\pi}{2}+2\pi n.\)
If having a negetive fraction at the begining of the answer looks weird, you can add \(2\pi\) to the negetive fraction to change the negetiive fraction to a positive fraction. When you do that to the answer to the first part of the equation, you get \(x=\frac{3\pi}{2}+2\pi n.\)
The answer to the second half of the equation is \(x=-\frac{\pi}{2}+2\pi n\) \((x=\frac{3\pi}{2}+2\pi n.).\)
