How can I rearrange this equation to make dx the focus?

[Tawaffles]

Hi,

First off, I am not doing a course in physics or maths so excuse me if this is a very basic question. I have the following equation (see attached) and I am trying to make dx the focus, ie:

dx = XYZ

Would anyone mind taking the time to help me do this?

Regards,

James

 

[Stephen Tashi]

Someone check this:

$$V_{C} = \frac{ V_{SE}}{2} ( 1 - \cos(x)) + \frac{V_{SC}}{2} ( 1 - \cos(x-dx)) + V_{DC}$$

$$V_{C} - V_{DC} - \frac{ V_{SE}}{2} (1 - \cos(x))) = \frac{V_{SC}}{2}(1 - \cos(x-dx))$$

$$\frac{2}{V_{SC}} ( V_{C} - V_{DC} - \frac{ V_{SE}}{2} (1 - \cos(x))) ) = 1 - \cos(x-dx)$$

$$\frac{2}{V_{SC}} ( V_{C} - V_{DC} - \frac{ V_{SE}}{2} (1 - \cos(x))) ) - 1 = - \cos(x-dx)$$

$$\cos(x-dx) = -\frac{2}{V_{SC}} ( V_{C} - V_{DC} - \frac{ V_{SE}}{2} (1 - \cos(x))) ) + 1$$

$$\cos(x-dx) = \frac{2}{V_{SC}} ( -V_{C} + V_{DC} +\frac{ V_{SE}}{2} (1 - \cos(x))) ) + 1$$

$$\cos(x-dx) = \frac{2}{V_{SC}} ( V_{DC} -+\frac{ V_{SE}}{2} (1 - \cos(x)))- V_{C} ) + 1$$

$$x - dx = \arccos( \frac{2}{V_{SC}} ( V_{DC} -+\frac{ V_{SE}}{2} (1 - \cos(x)))- V_{C} ) + 1 )$$

$$-dx = -x + \arccos( \frac{2}{V_{SC}} ( V_{DC} -+\frac{ V_{SE}}{2} (1 - \cos(x)))- V_{C} ) + 1 )$$

$$dx = x - \arccos( \frac{2}{V_{SC}} ( V_{DC} -+\frac{ V_{SE}}{2} (1 - \cos(x)))- V_{C} ) + 1 )$$