Solve for x

  • #1
bigpoppapump
7
0
having trouble with the following, if anyone could provide assistance it would be appreciated.

Solve for x:

1619175324265.png


and

Simplify the following:
1619175453718.png
 

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Answers and Replies

  • #2
jonah1
108
0
Beer soaked request follows.
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
Please show us what you have tried and exactly where you are stuck.

We can't help you if we don't where you are stuck.
 
  • #3
skeeter
1,104
1
https://mathhelpboards.com/attachments/1619175324265-png.11112/
change $\sin^2{x}$ to $(1-\cos^2{x})$ and solve the resulting quadratic equation for $\cos{x}$


https://mathhelpboards.com/attachments/1619175453718-png.11113/

change the cosecant and cotangent to factors in terms of sine & cosine, then simplify
 
  • #4
bigpoppapump
7
0
https://mathhelpboards.com/attachments/1619175324265-png.11112/
change $\sin^2{x}$ to $(1-\cos^2{x})$ and solve the resulting quadratic equation for $\cos{x}$


https://mathhelpboards.com/attachments/1619175453718-png.11113/

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.
 
  • #5
bigpoppapump
7
0
I have managed to solved these problems with confidence which is great. Thanks for your guidance.

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.
 
  • #6
skeeter
1,104
1
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.
 
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