Question

Find specific values between 0 and 2pi
tan^2(2x)-2tan(2x)-5
I already got the specific values but I need to understand how to get to them algebraically because I've just been graphing to find them

Answer 1

Hey Brill,\[\large\rm \left(\color{orangered}{\tan2x}\right)^2-2\color{orangered}{\tan2x}-5=0\]Let's make the substitution, \(\rm \color{orangered}{u=\tan2x}\)
giving us,\[\large\rm \left(\color{orangered}{u}\right)^2-2\color{orangered}{u}-5=0\]And from this point, we can solve it like a quadratic, ya?

Answer 2

i wonder if there is a type-o because there is no equation to solve @brill

Answer 3

he is right what you've given is a trigonometric expression not an equation.

Answer 4

Oh my bad I meant to put =0

Answer 5

And yeah @zepdrix I already solved it like a quadratic and got the values when I simplified them and then used tan inverse, but that would only give me one first quadrant angle when there are two in the first quadrant and 2 in the third quadrant

Answer 6

"Find specific values ... " of what? Continuous functions can and do take on infinite numbers of values within their respective ranges.

Answer 7

It's find specific values of x, as angle in radians, between 0 and 2pi

Answer 8

So, what constitutes a sample specific value? We have to know what we're looking for and how to recognize it when we think we've found it.

Answer 9

Umm not completely sure what you want as my answer but I got my final answer as
\[\frac{ \tan^{-1}( 1\pm \sqrt{6}) }{ 2 }\]
So that means that the number can be in all 4 quadrants because it is negative and positive
I also see that there are two answers per quadrant due to the period being modified in "tan(2x)"