Question

how do you solve the equation http://bamboodock.wacom.com/doodler/72ba5e33-0515-4e2f-9f92-2d5d7751ca80

Answer 1

I would call sin(2x) y
solve for y (you get a quadratic)

then solve for x using sin2x = y

Answer 2

what do you mean by sin(2x) y?

Answer 3

solve for y

\[ y^2 + (\frac{2-\sqrt{3}}{2})y -\frac{\sqrt{3}}{2} = 0\]

Answer 4

so y = sin2x?

Answer 5

yes, but first you need to find y

Answer 6

okay...im sorry im so lost here:(

Answer 7

is my way hard?

Answer 8

i want to figure it out with my way..if possible

Answer 9

I'll use y, just because it looks simpler

\[ y^2 - \frac{\sqrt{3}}{2} y + y - \frac{\sqrt{3}}{2} = 0 \]
reorder the terms
\[ y^2 + y- \frac{\sqrt{3}}{2} y - \frac{\sqrt{3}}{2} = 0 \]
put parens around each pair
\[ (y^2 + y)+ (-\frac{\sqrt{3}}{2} y - \frac{\sqrt{3}}{2}) = 0 \]
in the first pair factor out y. in the 2nd pair factor out -sqrt(3)/2
\[ y(y + 1)- \frac{\sqrt{3}}{2} (y +1) = 0 \]
now factor (y+1) from each term
\[ (y- \frac{\sqrt{3}}{2}) (y +1) = 0 \]
at this point you can say either
\[ (y- \frac{\sqrt{3}}{2}) = 0 \]
or
\[ (y +1)=0 \]
and you can find y
now sub in sin(2x) for y