# How to determine zero points of this function $2\cos(x) - 3\tan(x)=0$

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.

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I know I put the question there, but now, I need to determine zero points that this function makes. –  depecheSoul Feb 8 '14 at 16:45
What is $\pi/2$ doing here ? Could you explain ? –  Claude Leibovici Feb 8 '14 at 16:58
@ClaudeLeibovici cos function is periodic function. The graph of cos function meets x axis at PI/2, so and from there it hits x axis every PI –  depecheSoul Feb 8 '14 at 17:04

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.

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OK, i solved that, but the function is continus, so every some period of time, the function is again zero. I want ot determine, how big is the differecne between these two points –  depecheSoul Feb 8 '14 at 17:02
@depecheSoul See the added solution. –  Américo Tavares Feb 8 '14 at 17:09
Can you please explain to me how did you got to this solution. –  depecheSoul Feb 8 '14 at 17:13
@depecheSoul In the interval $[0,2\pi]$ $\sin (x)=1/2$ at $x=\pi/6$ and $x=\pi-\pi/6=5\pi/6$. The sinus function is periodic with period equal to $2\pi$. –  Américo Tavares Feb 8 '14 at 17:18
@depecheSoul Trying to address your doubt I've updated again my answer. –  Américo Tavares Feb 8 '14 at 18:29

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.

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