anonymous
  • anonymous
Solve the following equation, giving the exact solutions which lie in [0, 2π). (Enter your answers as a comma-separated list.) tan^2(x)=3/2sec(x)
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.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
can u plz rewrite ur question more explicitly...
anonymous
  • anonymous
\[\tan ^{2}(x)=\frac{ 3 }{ 2 }\sec(x)\]
anonymous
  • anonymous
what do you mean?

Looking for something else?

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

More answers

anonymous
  • anonymous
Change them into \(\sin\) and \(\cos\).
anonymous
  • anonymous
now its bit more clear
anonymous
  • anonymous
thus we have sin^2(x)/cos^2(x) =(3/2cosx) or (1 - cos^2(x))/cos^2(x) =3/(2cosx) ... i think now it can be solved after cross multiplication)
anonymous
  • anonymous
matri could you go on? Im kinda clueless
anonymous
  • anonymous
where do you find the prob in the above expression
anonymous
  • anonymous
how do i cross multiply that?
anonymous
  • anonymous
as u usually do it for regular equation
anonymous
  • anonymous
2cosx *((1 - cos^2(x)) = 3 cos^2(x) .....
anonymous
  • anonymous
2cosx *(1 - cos^2(x)) - 3 cos^2(x) =0 or cosx*(2(1 - cos^2(x)) -3cosx)=0...... cosx*(2(1 - cos^2(x) )-3cosx) =0 cosx*(2 - 2cos^2(x) -3cosx)=0 cosx*(2 -4cosx +cosx -2cos^2(x) )=0 cosx * (2*(1-2cosx) +cosx(1-2cosx))=0 or cosx *(2+cosx)*(1-2cosx)=0 hence cosx = 0 or cosx =1/2 [as cosx <>-2]
anonymous
  • anonymous
i hope further u can do it...

Looking for something else?

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