# TrigonometryFind x such that the given trigonometric expression equals zero.

#### anemone

##### MHB POTW Director
Staff member
Find [FONT=MathJax_Math]$$\displaystyle x$$ [/FONT]such that trigonometric $$\displaystyle \dfrac{\sin (3x) \cos (60^{\circ}-x)+1}{\sin (60^{\circ}-7x)-\cos (30^{\circ}+x)+m}=0$$ where $$\displaystyle m$$ is a fixed real number.

Hi all, I know the expression in the numerator has no real roots by checking it at W|A (plot the graph of y=sin(3x)cos(pi/3 -x)+1 - Wolfram|Alpha) and I daringly assumed the given expression can be rewritten in the form $$\displaystyle \dfrac{ab}{ac}=0$$ but, I failed to factor the numerator no matter how hard that I tried. I must be missing something very important here.

Thanks.

#### caffeinemachine

##### Well-known member
MHB Math Scholar
Find [FONT=MathJax_Math]$$\displaystyle x$$ [/FONT]such that trigonometric $$\displaystyle \dfrac{\sin (3x) \cos (60^{\circ}-x)+1}{\sin (60^{\circ}-7x)-\cos (30^{\circ}+x)+m}=0$$ where $$\displaystyle m$$ is a fixed real number.

Hi all, I know the expression in the numerator has no real roots by checking it at W|A (plot the graph of y=sin(3x)cos(pi/3 -x)+1 - Wolfram|Alpha) and I daringly assumed the given expression can be rewritten in the form $$\displaystyle \dfrac{ab}{ac}=0$$ but, I failed to factor the numerator no matter how hard that I tried. I must be missing something very important here.

Thanks.
I don't think there is a real solution to this. For if $x$ is a real number satisfying this then we must have $\sin(3x)\cos (60-x)=-1$.

Thus we either have:
$\sin(3x)=1, \cos(60-x)=-1$
Or we have:
$\sin(3x)=-1,\cos(60-x)=1$.

Both of these two cases fail to have a solution. (You can check this analytically, that is, you don't need a machine for this.)

#### HallsofIvy

##### Well-known member
MHB Math Helper
A fraction is 0 if and only if the numerator is 0 so "I know the expression in the numerator has no real roots" is all you need.

##### Well-known member
A fraction is 0 if and only if the numerator is 0 so "I know the expression in the numerator has no real roots" is all you need.
However the correct solution is missed. if we taken x = 30 ( assuming degrees)

sin 3x = - 1 and cos (60-x) = cos 90 = 1 we get

sin 3x cos(60-x) = - 1

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
However the correct solution is missed. if we taken x = 30 ( assuming degrees)

sin 3x = - 1 and cos (60-x) = cos 90 = 1 we get

sin 3x cos(60-x) = - 1
Not correct.

##### Well-known member
Not correct.
sorry. It was my mistake it should be -30 and not 30.

it was an oversite

then sin (- 90) = - 1 and cos (60-(-30)) = cos 90 = 1

Last edited:

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
sorry. It was my mistake it should be -30 and not 30.

it was an oversite

then sin (- 90) = - 1 and cos (60-(-30)) = cos 90 = 1
cos (90)=0 .

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
This can be seen by

$$\displaystyle \cos^2 x +\sin^2 x =1$$

The sum is always 1.

$$\displaystyle \cos^2 x +\sin^2 x =1$$