Thread: need help

1. need help

2.sinX - sin2X + cotg2X/2 = 10/3
2.sinX + sin2X

I have an exam tomorrow.... please help

2. Originally Posted by nezima
2.sinX - sin2X + cotg2X/2 = 10/3
2.sinX + sin2X
I will guess that "cotg" means "cotangent", that decimal points mean multiplication, and that 2s after function names indicate squaring of the function, so the equation is meant to be as follows:

. . . . . $\frac{\sin(x)\, -\, \sin^2(x)}{2\sin(x)\, +\, \sin^2(x)}\, +\, \cot^2\left(\frac{x}{2}\right)\, =\, \frac{10}{3}$

What were the instructions for this exercise? What have you tried so far? Where are you stuck?

Please be complete. Thank you!

3. $\frac{2\sin(x)\, -\, \sin(2x)}{2\sin(x)\, +\, \sin(2x)}\, +\, \cot^2\left(\frac{x}{2}\right)\, =\, \frac{10}{3}$ that is the exercise... x=? and.. sorry but i really cant explain you what i've tried .. in english ...

4. that is what i make.. and i cant find My mistake