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PreCal Help Please..I am so stuck on what to do

Mathematics
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Use the equation below to find the value of sin x. \[\frac{ \tan x }{ \cot x } - \frac{ \sec x }{ \cos x }= \frac{ 2 }{ \csc x }\]
write everything in terms of sin and cos
That is what I am not understanding. I have a really hard time writting out what it is supposed to be once you start changing everything.

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Other answers:

Hint: Write it this way. Then simply change everything to sin and cos \[\tan(x) \div \cot(x) - \sec(x) \div \cos(x) = 2 \div \csc(x)\]
So how would you change it?
\[\frac{\sin(x)}{\cos(x)} \div \frac{\cos(x)}{\sin(x)} - \frac{1}{\cos(x)} \div \cos(x) = 2 \div \frac{1}{\sin(x)}\]
so could the first two fracts cancel out?
Not exactly
How do you divide: \[\frac{1}{4} \div \frac{7}{6}\]
Why not? BTW i missed the lessons over this stuff so i am really lost on almost all of it.
Answer the previous question. How do you divide that?
Oh wait I remember
you switch the #'s of the second fraction so it is now 6/7 and you change divide to multiply. you now have 1/4 X 6/7 which is 6/28 simplified to 3/14
Yes, so do that same thing with the trig equation
so it is now \[\frac{ \sin^2x }{ \cos^2x }-\frac{ 1 }{ \cos x }\div \cos x=2\div \frac{ 1 }{ \sin x }\]??
Yes, but do the rest of them. Change all of the divisions into multiplication.
so it looks like\[\frac{ \sin^2x }{ \cos^2x }-\frac{ 1 }{ cosx }\times \frac{ cosx }{ 1 }=2\times \frac{ sinx }{ 1 }\]
Did I do that right?
If so, wouldn't 1/cosx times cosx/1 cancel out?
no, you did the cos(x)/1 incorrectly. cos(x) is already equal to cos(x)/1. You have to flip it.
oh duh ok so then it is \[\frac{ \sin^2 x}{ \cos^2x } -\frac{ 1 }{ \cos x }\times \frac{ 1 }{ \cos x }= 2\times \frac{ sinx }{ 1 }\]
Yes
so do the mult fractions combine to get -1/cos^2x
so then what do you do?
Yes, so what do you have after that?
\[\frac{ \sin^2x }{ \cos^2x }-\frac{ 1 }{ \cos^2x }= 2\times \frac{ sinx }{ 1 }\]
Okay, do you know how to combine fractions on the left side?
couldn't you turn 2x(sinx/1) into just 2sinx? and Im not sure what to do on the other side
What do I do next?
On the left side, you have to combine fractions if they have the same denominator.
For example: \[\frac{1}{4} + \frac{x}{4} = \frac{1 + x}{4}\]
The general rule for combining fractions with the same denominator is: \[\frac{a}{c} + \frac{b}{c} = \frac{a+b}{c}\]
so it is now \[\frac{ \sin^2x-1 }{ \cos^2x }=2sinx\]
or is the 2sin x supposed to be sin^2x
Very good
Now change the 1 to \(\sin^2x + \cos^2x\)
No, you have it right. 2sin(x)
wait, you put sin^2x + cos^2x instead of one? what is the new equation?
Yes, you put \(\sin^2x + \cos^2x\) in place of 1 because they are equal.
What do you get afterwards?
\[\frac{ \sin^2x-\sin^2x+\cos^2x }{ \cos^2x }=2sinx\] then you can subtract the sin^2x to get \[\frac{ \cos^2x }{ \cos^2x }=2sinx\] then if thats the case you would just have 2sinx
Actually, if you do it correctly you get: \[-\frac{ \cos^2x }{ \cos^2x }=2\sin x\]
Which simplifies to \[-1 = 2 \sin x\] Remember that we are still solving for x
so sinx is -2
You were supposed to get this after substituting the 1: \[\frac{ \sin^2x-(\sin^2x+\cos^2x) }{ \cos^2x }=2\sin x\] Then get : \[\frac{ \sin^2x-\sin^2x-cos^2x }{ \cos^2x }=2\sin x\]
or no sinx would be -1/2
Ultimately you end up isolating the sin(x) on the right side: \[-\frac{1}{2} = \sin x\]
Now all you have to do is take the inverse sin of both sides to get \[\sin^{-1}\left(-\frac{1}{2}\right) = x\]
So evaluate the left side using your calc. Let me know what you get. Also make sure it is in radian mode.
-.52
And thank you for your help :)
Well, in exact mode, you should have gotten \[x = -\frac{\pi}{6}\]
Nevertheless:\[-\frac{\pi}{6} \approx -.52\]

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