Exact solutions with radians

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!

Re: Exact solutions with radians

Whynot say $\displaystyle u = \cos a$ then solve $\displaystyle 2u^2+\sqrt{2}u = 0$

Re: Exact solutions with radians

So the answer would cos a = pi/4 ?

Or did i do something wrong.

Re: Exact solutions with radians

You're close, you should be solving $\displaystyle \cos a = 0, \frac{-1}{\sqrt{2}}$

Re: Exact solutions with radians

Dude, any chance you could show your working? Im confused as to how you got there.

Ill give you a foot rub :)

Re: Exact solutions with radians

Quote:

Originally Posted by

**FlexedCookie** Find all exact solutions in radians of

2cos^2 a + sqrt2 cos a = 0, when 0 _< a _< 2pi

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

factor ...

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

now set each factor equal to 0 and finish it.

Re: Exact solutions with radians

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...

Re: Exact solutions with radians

Quote:

Originally Posted by

**FlexedCookie** 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.

Re: Exact solutions with radians

Alrighty i asked but hejust gave me some answer that made even less sense.

any chance of your working mate? :( im desperate to figure out how this one works.

Re: Exact solutions with radians

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 $\displaystyle \begin{align*} 2x^2 + \sqrt{2}x = 0 \end{align*}$?

Re: Exact solutions with radians

I know how to solve the quadratic, its getting it to the quadratic im sucking at.

Re: Exact solutions with radians

Rightio. Managed to solve the first set of factors by setting it to 0. Got -1/sqrt2

Having trouble solving the second factor. i have no idea how to solve sqrtcos x +1 =0

any tips?