Question

solve the equation over the interval [0,360).
5cot^2theta-cot theta=2

Answer 1

\(\large\tt \begin{align} \color{black}{5cot^2\theta-cot \theta=2\\~\\
5cot^2\theta-cot \theta-2=0\\~\\
put~~cot \theta=x\\~\\
5x^2-x-2=0\\~\\
x=\dfrac{1\pm \sqrt{41}}{10}\\~\\
cot\theta=\dfrac{1\pm \sqrt{41}}{10}\\~\\
tan\theta=\dfrac{10}{1\pm \sqrt{41}}\\~\\
\theta=arctan(\dfrac{10}{1\pm \sqrt{41}})\\~\\
}\end{align}\)

Answer 2

no that makes no sense. how is that in the interval [0,360)?

Answer 3

i think u have to use calculator here,

Answer 4

http://www.wolframalpha.com/input/?i=tan%5Ctheta%3D%5Cdfrac%7B10%7D%7B1-+%5Csqrt%7B41%7D%7D+%2C0%3C%5Ctheta%3C360

Answer 5

click on the approximate forms

Answer 6

this is another
http://www.wolframalpha.com/input/?i=tan%5Ctheta%3D%5Cdfrac%7B10%7D%7B1%2B+%5Csqrt%7B41%7D%7D+%2C0%5Cleq+%5Ctheta%3C360

Answer 7

how are you getting x?

Answer 8

there u have to use quadratic formula

if
\(\Large ax^2+bx+c=0\)

then
\(\Large x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\)

Answer 9

ok thats where i was confused. thanks for your help