I need to determine zero point of this function.
$2\cos(x)-3\tan(x)=0$
I solved the equation, and the answer is 30° or $\pi/6$.
I know that zero pint of $\tan$ is $n\pi $, and for $\cos$ is $\frac{\pi}{2}+n\pi$.
I dont know how to put it all together.
Thanks.
I know that zero pint of $\tan$ is $n\pi $, and for $\cos$ is $\frac{\pi}{2}+n\pi$.
These facts alone are irrelevant for your question because on the left side of your equation you are summing the terms $2\cos x$ and $-3\tan x$. It would be very useful if you had a product of two factors, e.g.
$$(2\cos x)(-3\tan x)=0.$$
Then the zeros of this equation would be the zeros of $2\cos x=0$ and the zeros of $-3\tan x=0$.
Concerning your equation let us transform it into an equivalent one, e.g. as follows
\begin{eqnarray*} 2\cos x-3\tan x &=&0 \tag {1}\\ &\Leftrightarrow &2\cos x-3\frac{\sin x}{\cos x}=0,\qquad \cos x\neq 0 \\ &\Leftrightarrow &2\cos ^{2}x-3\sin x=0 \\ &\Leftrightarrow &2-2\sin ^{2}x-3\sin x=0.\tag {2} \end{eqnarray*}
Let $y=\sin x$. Then \begin{equation*} 2-2y^{2}-3y=0\Leftrightarrow y=-2,y=\frac{1}{2}.\tag {3} \end{equation*}
Since the solution $y=\sin x=-2$ is impossible, the problem is reduced to finding the roots of \begin{equation*} \sin x=\frac{1}{2}.\tag {4} \end{equation*}
The general solution of $(4) $ is $x=\frac{\pi }{6}+2k\pi$ and $x=\frac{5\pi }{6}+2k\pi$, with $k\in \mathbb{Z}$, because, in the interval $[0,2\pi]$, $\sin (x)=1/2$ at $x=\pi/6$ and $x=\pi-\pi/6=5\pi/6$, and not only the point $ x=\pi/6 $, as found by you, (see Wikipedia entry Trigonometric functions on the unit circle) and the sinus function is periodic with period equal to $2\pi$.
Alternatively look at the graph of $y=\sin(x)$ for $-2\pi\le x\le 2\pi$, and the horizontal line $y=1/2$, which crosses $ y=\sin (x) $ at the points mentioned above, and you can see the period of this trigonometric function.
Set $u = \cos \theta$, this gives: $$ 2 u - 3 \frac{\sqrt{1 - u^2}}{u} = 0 \\ 2 u^2 = 3 \sqrt{1 - u^2} \\ 4 u^4 + 9 u^2 - 9 = 0 $$ The last one is a quadratic in $u^2$. Just be careful that the squaring didn't introduce spurious solutions.
