Question

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

Answer 1

Use the equation below to find the value of sin x.
\[\frac{ \tan x }{ \cot x } - \frac{ \sec x }{ \cos x }= \frac{ 2 }{ \csc x }\]

Answer 2

write everything in terms of sin and cos

Answer 3

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.

Answer 4

Hint: Write it this way. Then simply change everything to sin and cos

\[\tan(x) \div \cot(x) - \sec(x) \div \cos(x) = 2 \div \csc(x)\]

Answer 5

So how would you change it?

Answer 6

\[\frac{\sin(x)}{\cos(x)} \div \frac{\cos(x)}{\sin(x)} - \frac{1}{\cos(x)} \div \cos(x) = 2 \div \frac{1}{\sin(x)}\]

Answer 7

so could the first two fracts cancel out?

Answer 8

Not exactly

Answer 9

How do you divide:

\[\frac{1}{4} \div \frac{7}{6}\]

Answer 10

Why not? BTW i missed the lessons over this stuff so i am really lost on almost all of it.

Answer 11

Answer the previous question. How do you divide that?

Answer 12

Oh wait I remember

Answer 13

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

Answer 14

Yes, so do that same thing with the trig equation

Answer 15

so it is now \[\frac{ \sin^2x }{ \cos^2x }-\frac{ 1 }{ \cos x }\div \cos x=2\div \frac{ 1 }{ \sin x }\]??

Answer 16

Yes, but do the rest of them. Change all of the divisions into multiplication.

Answer 17

so it looks like\[\frac{ \sin^2x }{ \cos^2x }-\frac{ 1 }{ cosx }\times \frac{ cosx }{ 1 }=2\times \frac{ sinx }{ 1 }\]

Answer 18

Did I do that right?

Answer 19

If so, wouldn't 1/cosx times cosx/1 cancel out?

Answer 20

no, you did the cos(x)/1 incorrectly. cos(x) is already equal to cos(x)/1. You have to flip it.

Answer 21

oh duh ok so then it is \[\frac{ \sin^2 x}{ \cos^2x } -\frac{ 1 }{ \cos x }\times \frac{ 1 }{ \cos x }= 2\times \frac{ sinx }{ 1 }\]

Answer 22

Yes

Answer 23

so do the mult fractions combine to get -1/cos^2x

Answer 24

so then what do you do?

Answer 25

Yes, so what do you have after that?

Answer 26

\[\frac{ \sin^2x }{ \cos^2x }-\frac{ 1 }{ \cos^2x }= 2\times \frac{ sinx }{ 1 }\]

Answer 27

Okay, do you know how to combine fractions on the left side?

Answer 28

couldn't you turn 2x(sinx/1) into just 2sinx? and Im not sure what to do on the other side

Answer 29

What do I do next?

Answer 30

On the left side, you have to combine fractions if they have the same denominator.

Answer 31

For example:
\[\frac{1}{4} + \frac{x}{4} = \frac{1 + x}{4}\]

Answer 32

The general rule for combining fractions with the same denominator is:

\[\frac{a}{c} + \frac{b}{c} = \frac{a+b}{c}\]

Answer 33

so it is now \[\frac{ \sin^2x-1 }{ \cos^2x }=2sinx\]

Answer 34

or is the 2sin x supposed to be sin^2x

Answer 35

Very good

Answer 36

Now change the 1 to \(\sin^2x + \cos^2x\)

Answer 37

No, you have it right. 2sin(x)

Answer 38

wait, you put sin^2x + cos^2x instead of one? what is the new equation?

Answer 39

Yes, you put \(\sin^2x + \cos^2x\) in place of 1 because they are equal.

Answer 40

What do you get afterwards?

Answer 41

\[\frac{ \sin^2x-\sin^2x+\cos^2x }{ \cos^2x }=2sinx\]
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

Answer 42

Actually, if you do it correctly you get:

\[-\frac{ \cos^2x }{ \cos^2x }=2\sin x\]

Answer 43

Which simplifies to

\[-1 = 2 \sin x\]

Remember that we are still solving for x

Answer 44

so sinx is -2

Answer 45

You were supposed to get this after substituting the 1:


\[\frac{ \sin^2x-(\sin^2x+\cos^2x) }{ \cos^2x }=2\sin x\]

Then get :
\[\frac{ \sin^2x-\sin^2x-cos^2x }{ \cos^2x }=2\sin x\]

Answer 46

or no sinx would be -1/2

Answer 47

Ultimately you end up isolating the sin(x) on the right side:

\[-\frac{1}{2} = \sin x\]

Answer 48

Now all you have to do is take the inverse sin of both sides to get

\[\sin^{-1}\left(-\frac{1}{2}\right) = x\]

Answer 49

So evaluate the left side using your calc. Let me know what you get. Also make sure it is in radian mode.

Answer 50

-.52

Answer 51

And thank you for your help :)

Answer 52

Well, in exact mode, you should have gotten \[x = -\frac{\pi}{6}\]

Answer 53

Nevertheless:\[-\frac{\pi}{6} \approx -.52\]