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

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,756
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.

Could anyone please help me out with this problem?

Thanks.
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
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.

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.)
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
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.
 

kaliprasad

Well-known member
Mar 31, 2013
1,322
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
Jan 17, 2013
1,667
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.
 

kaliprasad

Well-known member
Mar 31, 2013
1,322
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
Jan 17, 2013
1,667
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
Jan 17, 2013
1,667
This can be seen by

\(\displaystyle \cos^2 x +\sin^2 x =1 \)

The sum is always 1.
 

kaliprasad

Well-known member
Mar 31, 2013
1,322