1. ## Algebra

Solve the equation:

$\displaystyle cos x+\sqrt{1-\tfrac{1}{2} sin2x} = 0$, in the interval $\displaystyle 0^\circ \le x < 360^\circ$

Thanks!

2. Originally Posted by rednest
Solve the equation:

$\displaystyle cos x+\sqrt{1-\tfrac{1}{2} sin2x} = 0$, in the interval $\displaystyle 0^\circ \le x < 360^\circ$

Thanks!
$\displaystyle \sqrt{1-\tfrac{1}{2} \sin 2x} = -\cos x$

3. Here's what I did. I'm pretty sure it's right:

We have-

$\displaystyle \cos{x} + \sqrt{1 - \frac{1}{2}\sin{2x}} = 0$

First:

$\displaystyle \sqrt{1 - \frac{1}{2}\sin{2x}} = -\cos{x}$

Next:

$\displaystyle 1 - \frac{1}{2}\sin{2x} = \cos^2{x}$

Keep in mind that $\displaystyle (-\cos{x})^2 = \cos^2{x}$

Then:

$\displaystyle -\frac{1}{2}\sin{2x} = \cos^2{x} - 1$

Now we remember that:

$\displaystyle \sin^2{x} + cos^2{x} = 1$

Therefore:

$\displaystyle \cos^2{x} - 1 = -\sin^2{x}$

So:

$\displaystyle -\frac{1}{2}\sin{2x} = -\sin^2{x}$

Negatives can cancel, giving us:

$\displaystyle \frac{1}{2}\sin{2x} = \sin^2{x}$

Now, we go to another identity:

$\displaystyle \sin{2x} = 2\sin{x}\cos{x}$

So:

$\displaystyle \sin{x}\cos{x} = \sin^2{x}$

$\displaystyle \left(\frac{1}{2}(2\sin{x}\cos{x}) = \sin{x}\cos{x}\right)$

Now:

$\displaystyle \sin{x} = \sin{x}\tan{x}$

And finally:

$\displaystyle 1 = \tan{x}$

$\displaystyle x = 45^{\circ}, \ 225^{\circ}$

And there you go.

4. Originally Posted by Aryth

Now:

$\displaystyle \sin{x} = \sin{x}\tan{x}$

And finally:

$\displaystyle 1 = \tan{x}$

$\displaystyle x = 45^{\circ}, \ 225^{\circ}$

And there you go.
we cannot do this immediately, since $\displaystyle x$ can be zero (and pi) and so does $\displaystyle \sin x$..

so, the best way is to do this: $\displaystyle \sin{x}\tan{x} - \sin x = (\sin x) (\tan x - 1) = 0$..

and there you go.. $\displaystyle 1 = \tan{x}$ and also $\displaystyle \sin x = 0$

after finding the values of $\displaystyle x$, try to check whether it fits the original equation..