Solve the equation:
$\displaystyle cos x+\sqrt{1-\tfrac{1}{2} sin2x} = 0$, in the interval $\displaystyle 0^\circ \le x < 360^\circ$
Please help me on how to simplify this equation in the exam (any techniques?).
Thanks!
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.
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..