Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I need to determine zero point of this function.


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.


share|cite|improve this question
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.

enter image description here

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.