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I tried factoring but I couldn't do it.

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Hint: Quadratic formula--$$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ are the roots of $ax^2+bx+c=0$($a \neq 0$) – user21436 Apr 6 '12 at 17:20
I tied that but there were answers that were too complex (i think): – David Apr 6 '12 at 17:22
You had a + in place of minus. See this for instance. – user21436 Apr 6 '12 at 17:24
Notice that 360^\circ looks like this: $360^\circ$. But 360^o looks like this: $360^o$. The "$o$" gets italicized. (I edited the question accordingly.) – Michael Hardy Apr 6 '12 at 17:32
Thanks for the edit! :) – David Apr 6 '12 at 17:41
up vote 1 down vote accepted

Let $w=\sin x$, and consider the equation $$3w^2+3w-2=0.$$ The polynomial of the left does not factor over the integers. To find the roots, the simplest approach is probably to use the Quadratic Formula. Recall that the solutions of the equation $aw^2+bw+c=0$, where $a \ne 0$, are given by $$w=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$ So the solutions of your equation are given by $$w=\frac{-3\pm\sqrt{33}}{6}.$$ Now it continues much like your earlier question.

Note that the solution $w=\frac{-3-\sqrt{33}}{6}$ is less than $-1$, and therefore cannot be the sine of anything. The other solution is between $0$ and $1$. There will be two values of $x$ in your interval whose sine is $\frac{-3+\sqrt{33}}{6}$. One of them will be directly given to you by your calculator, and will be between $0$ and $90$. Call it $x_1$. The other value of $x$ will be $180-x_1$. (Look at a graph of the sine curve to see that there are two values of $x$, one in the first quadrant and the other in the second quadrant.)

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To the OP: Note that one of the solutions is not a valid solution. (Why?) – user21436 Apr 6 '12 at 17:28

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