MHB Find x such that the given trigonometric expression equals zero.

AI Thread Summary
The discussion centers on solving the trigonometric equation involving the expression $$\dfrac{\sin (3x) \cos (60^{\circ}-x)+1}{\sin (60^{\circ}-7x)-\cos (30^{\circ}+x)+m}=0$$ where $$m$$ is a fixed real number. Participants note that the numerator has no real roots, as confirmed by graphing tools. One contributor suggests that for the expression to equal zero, the numerator must equal zero, leading to the conclusion that there are no real solutions. However, a mistake is later acknowledged regarding the value of $$x$$, with a correction indicating that $$x = -30$$ yields valid results. The discussion concludes with an acknowledgment of the oversight in the initial analysis.
anemone
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Find [FONT=MathJax_Math]$$x$$ such that trigonometric $$\dfrac{\sin (3x) \cos (60^{\circ}-x)+1}{\sin (60^{\circ}-7x)-\cos (30^{\circ}+x)+m}=0$$ where $$m$$ 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 $$\dfrac{ab}{ac}=0$$ but, I failed to factor the numerator no matter how hard that I tried. I must be missing something very important here.

Could anyone please help me out with this problem?

Thanks.
 
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anemone said:
Find [FONT=MathJax_Math]$$x$$ such that trigonometric $$\dfrac{\sin (3x) \cos (60^{\circ}-x)+1}{\sin (60^{\circ}-7x)-\cos (30^{\circ}+x)+m}=0$$ where $$m$$ 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 $$\dfrac{ab}{ac}=0$$ but, I failed to factor the numerator no matter how hard that I tried. I must be missing something very important here.

Could anyone please help me out with this problem?

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.)
 
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.
 
HallsofIvy said:
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
 
kaliprasad said:
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.
 
ZaidAlyafey said:
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:
kaliprasad said:
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 .
 
This can be seen by

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

The sum is always 1.
 
ZaidAlyafey said:
This can be seen by

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

The sum is always 1.
I realize

it was my goofup
thanks
 
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