• Mar 8th 2012, 06:15 PM
Im struggling to get my head around radians, and with this question im lost.

And if you could explain what you're oing as wel so i can follow on thatd be brilliant.

Find all exact solutions in radians of
2cos^2 a + sqrt2 cos a = 0, when 0 _< a _< 2pi

_< is greater or equal to.

Fanks!
• Mar 8th 2012, 06:18 PM
pickslides
Whynot say $u = \cos a$ then solve $2u^2+\sqrt{2}u = 0$
• Mar 9th 2012, 02:38 PM
So the answer would cos a = pi/4 ?
Or did i do something wrong.
• Mar 9th 2012, 04:30 PM
pickslides
You're close, you should be solving $\cos a = 0, \frac{-1}{\sqrt{2}}$
• Mar 10th 2012, 09:30 PM
Dude, any chance you could show your working? Im confused as to how you got there.
Ill give you a foot rub :)
• Mar 11th 2012, 05:25 AM
skeeter
Quote:

Find all exact solutions in radians of
2cos^2 a + sqrt2 cos a = 0, when 0 _< a _< 2pi

$2\cos^2{a} + \sqrt{2} \cos{a} = 0$

factor ...

$\sqrt{2} \cos{a}(\sqrt{2} \cos{a} + 1) = 0$

now set each factor equal to 0 and finish it.
• Mar 11th 2012, 06:35 PM
Im so sorry but that just made me even more confused :( mind putting in your working? Im desperately trying to get my head around how these things are working...
• Mar 12th 2012, 07:18 AM
skeeter
Quote:

Im so sorry but that just made me even more confused :( mind putting in your working? Im desperately trying to get my head around how these things are working...

What do you know about factoring and using the zero product property to solve equations? If you are unfamiliar with this procedure, then I recommend you speak to your instructor.
• Mar 14th 2012, 06:23 PM

any chance of your working mate? :( im desperate to figure out how this one works.
• Mar 14th 2012, 07:27 PM
Prove It
Can you solve a general quadratic? I.e. do you know that if \displaystyle \begin{align*} ax^2 + bx + c = 0 \end{align*} then \displaystyle \begin{align*} x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \end{align*}?

If so, can you solve \displaystyle \begin{align*} 2x^2 + \sqrt{2}x = 0 \end{align*}?
• Mar 14th 2012, 09:47 PM