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#### bigpoppapump

##### New member

- Apr 23, 2021

- 7

- Thread starter bigpoppapump
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- Apr 23, 2021

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- Feb 21, 2015

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Please show us what you have tried and exactly where you are stuck.having trouble with the following, if anyone could provide assistance it would be appreciated.

Solve for x:

View attachment 11112

and

Simplify the following:

View attachment 11113

We can't help you if we don't where you are stuck.

- Mar 1, 2012

- 871

change $\sin^2{x}$ to $(1-\cos^2{x})$ and solve the resulting quadratic equation for $\cos{x}$

change the cosecant and cotangent to factors in terms of sine & cosine, then simplify

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- Apr 23, 2021

- 7

change $\sin^2{x}$ to $(1-\cos^2{x})$ and solve the resulting quadratic equation for $\cos{x}$

change the cosecant and cotangent to factors in terms of sine & cosine, then simplify

Thank you. This helps, I was stuck but I have a good idea on how to solve both of these. Will work on it tonight.

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- #5

- Apr 23, 2021

- 7

I have a word problem that I’m finding it difficult to convert into an equation. Could some direction be given so I can then run with it and complete.

The question is...

An electrical circuit runs at 50Hz at 0.5amps. Due to a lag in the switch, the first maximum current is reached at 6milliseconds. Assuming no variation, find an equation to model the current in this circuit using time in milliseconds.

- Mar 1, 2012

- 871

frequency is the reciprocal of period (time to complete one cycle of AC)

$T = \dfrac{1}{50} = 0.02 \text{ sec } = 20 \text{ milliseconds}$

current flow (with no lag) as a function of time in milliseconds ...

$A = 0.5 \sin\left(\dfrac{\pi}{10} \cdot t \right)$

For that period, the sinusoidal graph of current would peak at $\dfrac{T}{4} = 5 \text{ milliseconds}$

Due to the lag, there is a 1 millisecond horizontal shift in the graph ...

**In future, please start a new problem with a new thread.**

$T = \dfrac{1}{50} = 0.02 \text{ sec } = 20 \text{ milliseconds}$

current flow (with no lag) as a function of time in milliseconds ...

$A = 0.5 \sin\left(\dfrac{\pi}{10} \cdot t \right)$

For that period, the sinusoidal graph of current would peak at $\dfrac{T}{4} = 5 \text{ milliseconds}$

Due to the lag, there is a 1 millisecond horizontal shift in the graph ...

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