# cosx(2sinx + sqrt3)(-sqrt2 cosx + 1) = 0??

Jul 25, 2016

Maybe you meant "how do I solve this equation". By the zero-product property, one or all of these factors can be equal to $0$.

$\cos x \left(2 \sin x + \sqrt{3}\right) \left(- \sqrt{2} \cos x + 1\right) = 0$

• When $\cos x = 0$, $x = \frac{\pi}{2} , \frac{3 \pi}{2} , . . .$, or:

$\textcolor{b l u e}{x = \frac{\pi}{2} \pm n \pi}$

• When $2 \sin x + \sqrt{3} = 0$, $\sin x = - \frac{\sqrt{3}}{2}$. That can occur for ${240}^{\circ}$, ${600}^{\circ}$, ${960}^{\circ}$, etc., and ${300}^{\circ}$, ${660}^{\circ}$, ${1020}^{\circ}$, etc. Or:

$\textcolor{b l u e}{x = \frac{4 \pi}{3} \pm 2 n \pi}$, where $n$ is an integer.
$\textcolor{b l u e}{x = \frac{5 \pi}{3} \pm 2 n \pi}$, where $n$ is an integer.

• When $1 - \sqrt{2} \cos x = 0$, $\cos x = \frac{1}{\sqrt{2}} \cdot \left(\frac{\sqrt{2}}{\sqrt{2}}\right) = \frac{\sqrt{2}}{2}$. This occurs when $x = {45}^{\circ} , {315}^{\circ} , {405}^{\circ} , {675}^{\circ} , . . .$. So $x = \frac{\pi}{4} , \frac{7 \pi}{4} , \frac{9 \pi}{4} , \frac{15 \pi}{4} , . . .$, or:

$\textcolor{b l u e}{x = 2 n \pi \pm \frac{\pi}{4}}$, where $n$ is an integer.

Note that written this way, this already includes both $\frac{\pi}{4}$ and $\frac{7 \pi}{4}$ for $0 < x < 2 \pi$ and other corresponding angles in subsequent cycles. For example:

$n = - 1 \to x = - \frac{9 \pi}{4} , - \frac{7 \pi}{4}$
$n = 0 \to x = - \frac{\pi}{4} , \frac{\pi}{4}$
$n = 1 \to x = \frac{7 \pi}{4} , \frac{9 \pi}{4}$
$n = 2 \to x = \frac{15 \pi}{4} , \frac{17 \pi}{4}$

If I took this at face value, I honestly don't think this is equal to $0$.

http://www.wolframalpha.com/input/?i=Is+cosx%282sinx%2Bsqrt3%29%28+-sqrt2cosx%2B1%29+%3D+0

If we tried, we would get:

cosx(2sinx+sqrt3)( -sqrt2cosx+1) stackrel(?)(=) 0

(2sinxcosx+sqrt3cosx)( -sqrt2cosx+1) stackrel(?)(=) 0

-2sqrt2sinxcos^2x + 2sinxcosx - sqrt6cos^2x + sqrt3cosx stackrel(?)(=) 0

2sinxcosx + sqrt3cosx stackrel(?)(=) 2sqrt2sinxcos^2x + sqrt6cos^2x

(2sinx + sqrt3)cosx stackrel(?)(=) (2sqrt2sinx + sqrt6)cos^2x

cancel((2sinx + sqrt3))cosx stackrel(?)(=) sqrt2cancel((2sinx + sqrt3))cos^2x

cosx stackrel(?)(=) sqrt2cos^2x

1 stackrel(?)(=) sqrt2cosx

$\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \ne \cos x$ necessarily. Functions are not always the same as constants.

$\therefore$ with no value for $x$, this is an impossible proof.