anonymous
  • anonymous
tan^2 x + sqr3 tanx = 0 solve please
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
tan x = sin x / cos x In each quadrant, tan takes on all the values of one sign with magnitudes ranging from 0 to ∞. It is negative in [π/2, π] and in [3π/2, 2π]. Knowing that tan π/6 = tan 30º = (sin π/6)/(cos π/6) = (1/2)(√3/2) = 1/√3, and that tan(x+π/2) = -1/tan x, you can see that tan(2π/3) = tan(π/6 + π/2) = -√3, and that tan(x+π) = tan(x+π/2+π/2) = -1/tan(x+π/2) = tan x, so tan(5π/3) = tan(2π/3) = -√3 so x = 2π/3, 5π/3, if you're in the [0 to 2π] full circle; x = 2π/3, -π/3, if you're in the [-π to π] full circle Lemme know if this helps, if you need more help @~Chance~
Psymon
  • Psymon
You can just factor this. You would solve it the same way you sould solve \[x ^{2}+\sqrt{3}x \implies x(x+\sqrt{3})\] Just do that, but factor out a tan(x) instead. Then you set both factors equal to 0 and solve for each.
anonymous
  • anonymous
thanks both of you

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Best responce me please?

Looking for something else?

Not the answer you are looking for? Search for more explanations.