Directions: Find all solutions in the interval (0, 2pi] Problem: tan^2(x)*csc(x) = tan^2(x) Translated into English since it might look confusing typed out... the tangeant squared of x times the cosecant of x equals the tangeant squared of x. That is not the tangeant of (x^2). I determined there were no solutions, but that wasn't the right answer. I still think I got it right. If I can get the points for this one problem, I will have an A instead of a B in that class for this past Spring semester. Can you guys tell me what answer you get and how you worked it out?

the thing that confuses me is csc(x), I´m only familiar with tan(x), cos(x) and sin(x) or perhaps we write it differnently, but I doubt that. well, physics final tomorrow... I´d better go study

jakeman just spelled it wrong ...I would have been able to do that problem for you a LONG time ago, but I forgot all my trig..isn't tangeant, cosxsin or something like that? I seem to remember you have to turn them all into like terms or something...but my brain explodes when thinking that far back ('94 guys, almost 10 years.....ouch!!!! *bonk*

Well since the first function times another function equals the first function, the second function must be 1. tan^2(x)*csc(x)=tan^2(x) -I know this isn't technically allowed, but divide both sides by tan^2(x) csc(x)=1 so 0 and 2pi edit- 0 and 2pi is wrong, I mixed up csc with sec...it's pi/2 edit2-oh yeah, tan is undefined with that. There's no answer.

From that Chen guys post in the other forum "sin(x) = sin^2(x) Or: sin(x) * [1 - sin(x)] = 0" wouldn´t it be sin(x) = 1- cos^2(x)

That's the exact logic I used. I got to an answer of pi/2, but that made tan undefined so I said there was no answer. But I neglected to consider the possible solution of tan(x) equalling 0. At 0 and pi, the equation equates to 0 * undefined = 0 . I guess "undefined" is legit when it's multiplied by 0.

well if its basicaly a*b=a couldnt you work your way backwards using the negative function? though im not familiar with cosecant. and i cant find a button familiar to csc on ti-83.. Is this some weird form of cosign?

csc = 1/sin So calculate the sin, then hit your inverse key. Or store the value, clear, 1 divided by *recall value*, enter.

It's the sin^-1 key on your calculator. Consider replacing tangent in the problem with its definition (sin / cos). My solution is X = π / 2

The tangeant is undefined at pi/2 sin(pi/2) = 1 cos(pi/2) = 0 tan(x) = sin(x)/cos(x) tan(pi/2) = sin(pi/2)/cos(pi/2) tan(pi/2) = 1/0 tan(pi/2) = undefined Be careful when multiplying or dividing both sides of an equation by a variable quantity. You can introduce extraneous solutions that must be checked in the original equation at the end.

Not quite related to the problem, but I have my own problem, i need to find the derivitive of y = e^e^x

ln both sides to get lny = lne^e^x lne = 1 so lny = e^x dy/dx (lny) = dy/dx (e^x) dy/dx (1/y) = e^x NIGGA PLZ

Ask your prof, I don't see where that could come from unless they provide you with conditions to solve.

Lolz. Nevermind. They're not solving with ln, just solving with basics. So you have y = e^(e^x) Power rule, multiply the original part by the derivative of the exponent. The derivative of e^x is e^x so just multiple e^e^x by e^x and you end up with y = (e^x)(e^e^x) Simplify y = e^(x+e^x) by exponent laws. Seems weird to me.