Have no concept of where to go from here...

[jtw2e2]

Find all solutions of the equation: 5tan(x)=15

Obviously tan(x)=3

Using a right triangle, I found:
sin(x) = 3/?10
cos(x) = 1/?10

But I'm stumped on where to go from there. Would someone kindly direct me to how I might go about solving for x for the above equation?

 

[Deleted member 4993]

Find all solutions of the equation: 5tan(x)=15

Obviously tan(x)=3

Using a right triangle, I found:
sin(x) = 3/?10
cos(x) = 1/?10

But I'm stumped on where to go from there. Would someone kindly direct me to how I might go about solving for x for the above equation?

Are you allowed to use calculator?

Is the domain bounded - like 0 ? x ? ?/2 ?

 

[jtw2e2]

Find all solutions of the equation: 5tan(x)=15

Obviously tan(x)=3

Using a right triangle, I found:
sin(x) = 3/?10
cos(x) = 1/?10

But I'm stumped on where to go from there. Would someone kindly direct me to how I might go about solving for x for the above equation?

Are you allowed to use calculator?

Is the domain bounded - like 0 ? x ? ?/2 ?

From my understanding, yes we are restricting the domain to one period of the trig function. However, that was not mentioned in this problem. Generally we have not been allowed to use calculators, although we have no direction on this particular assignment. It is web-based homework and the 2nd part of the problem does call for a decimal answer to 5 places, so I'm assuming a calculator is appropriate. But at this point, I wouldn't know how to use it to solve this. If it were an easy tangent, for example, tan(x)=1 then obviously x would have to equal pi/4. But I'm at a loss for how to find the value of x when it is not one of the points we've memorized on the unit circle.

 

[Deleted member 4993]

If you have a decent (> $10) Scientific Calculator - it should have buttons like sin[sup:1yxsxcgx]-1[/sup:1yxsxcgx], cos[sup:1yxsxcgx]-1[/sup:1yxsxcgx] and tan[sup:1yxsxcgx]-1[/sup:1yxsxcgx].

Use the user's manual to learn to use those buttons.

 

[jtw2e2]

If you have a decent (> $10) Scientific Calculator - it should have buttons like sin[sup:36nmuax6]-1[/sup:36nmuax6], cos[sup:36nmuax6]-1[/sup:36nmuax6] and tan[sup:36nmuax6]-1[/sup:36nmuax6].

Use the user's manual to learn to use those buttons.

I do have a TI-84plus but not with me. I was hoping to use the arctan function of google calculator but didn't know what to type in.

 

[Deleted member 4993]

If you have a decent (> $10) Scientific Calculator - it should have buttons like sin[sup:2prod3x1]-1[/sup:2prod3x1], cos[sup:2prod3x1]-1[/sup:2prod3x1] and tan[sup:2prod3x1]-1[/sup:2prod3x1].

Use the user's manual to learn to use those buttons.

I do have a TI-84plus but not with me. I was hoping to use the arctan function of google calculator but didn't know what to type in.

Why not arctan(3) - if you are wrong, it won't bite you!

Remember the answer comes back in "radians".

 

[jtw2e2]

[quote="Subhotosh Khan":1h9ikx3b]If you have a decent (> $10) Scientific Calculator - it should have buttons like sin[sup:1h9ikx3b]-1[/sup:1h9ikx3b], cos[sup:1h9ikx3b]-1[/sup:1h9ikx3b] and tan[sup:1h9ikx3b]-1[/sup:1h9ikx3b].

Use the user's manual to learn to use those buttons.

I do have a TI-84plus but not with me. I was hoping to use the arctan function of google calculator but didn't know what to type in.

Why not arctan(3) - if you are wrong, it won't bite you!

Remember the answer comes back in "radians".[/quote:1h9ikx3b]

I got 1.24904577 rad. According to my homework's website, the correct answer is 1.37340 + pi k for (smallest ?-value) and "NONE" for (largest ?-value). Don't know where I went wrong.

 

[Deleted member 4993]

I got 1.24904577 rad. According to my homework's website, the correct answer is 1.37340 + pi k for (smallest ?-value) and "NONE" for (largest ?-value). Don't know where I went wrong.

[/quote][/quote]

If you posted your problem correctly - the answer you got is correct.

 

[jtw2e2]

If you posted your problem correctly - the answer you got is correct.

You're right. It took that value this time. Thanks for the help. I'm sure I'll be back for more within the hour. :lol: