Question

Solve each equation.

Answer 1

sin^2theta + cos(theta) = 2
(Hint: Use the Pythagorean identity sin^2theta _ cos^2 theat = 1 to replace sin^2 theta in the given equation.)

Answer 2

I replaced sin^2 theta with (1 - cos^2 theta) but then I got stuck...
(1 - cos^2theta) + costheta = 2
Should I nudge the 1 to the other side and factor out a costheta? Or pull everything on one side and use the quadratic formula?

Answer 3

Get everything on one side.

You're on the right track with the quad. formula, but try factoring it first.

Answer 4

\[\large (1 - \cos^2\theta) + \cos \theta = 2 \] \[\large \cos^2 \theta - \cos \theta +1 = 0\]So it doesn't factor, use the quadratic formula.

Answer 5

Okay, I'll let you know what I get.

Answer 6

Oh boy... this is a hairy one.

Answer 7

\[-\cos \theta \pm \sqrt{\cos^1 \theta - 4\cos^2 \theta}\]
All of that ^ over this:
\[-2\cos^2 \theta\]

Answer 8

Now... I feel like a cos theta can be factored out of the square root? (it is supposed by be cos^2, not to the first power btw)

Answer 9

Woahhh you're way overcomplicating this. All you need is the coefficients, not the cos theta's
\[\large \cos^2 \theta - \cos \theta +1 = 0\]
It's the same thing as \[\large x^2 - x +1 = 0\]
The only difference is that after you use the quadratic formula, you need to remember it's cos theta equal to w/e solutions you find, not x.

Answer 10

Haha. :P
Lemme retry.

Answer 11

Does this look right?
\[\frac{ 1 }{ 2 } \pm -\sqrt{3}/2\]

Answer 12

So (1 + sqrt3)/2 and (1 - sqrt3)/2
hmmmm I don't know.

Answer 13

\[\large x^2 - x +1 = 0 \] \[\large x = \frac{ 1 \pm \sqrt{1- 4*1*1} }{ 2 }\]
Look closely at the square root... :P

Answer 14

Negative on the inside, aha.

Answer 15

And if you look at the original equation... the only way it's true is if sin theta AND cos theta are both 1

and they never BOTH equal 1 (for the same value of theta), it's only ever one or the other.

Answer 16

http://www.wolframalpha.com/input/?i=sin%5E2theta+%2B+cos(theta)+%3D+2

See the graph, no intersections.

Answer 17

Would I say no solutions or undefined?

Answer 18

Undefined doesn't really make sense when trying to solve an equation. It only makes sense for something like the value of \[\arccos 2\]

Answer 19

Right.
Thanks a bunch, your help has been invaluable!

Answer 20

Welcome!