1. ## Algebra

Solve the equation:

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

Thanks!

2. Originally Posted by rednest
Solve the equation:

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

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

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

We have-

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

First:

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

Next:

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

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

Then:

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

Now we remember that:

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

Therefore:

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

So:

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

Negatives can cancel, giving us:

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

Now, we go to another identity:

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

So:

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

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

Now:

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

And finally:

$1 = \tan{x}$

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

And there you go.

4. Originally Posted by Aryth

Now:

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

And finally:

$1 = \tan{x}$

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

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

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

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

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