Find x such that the given trigonometric expression equals zero.

[anemone]

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]

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]

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]

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

 

[alyafey22]

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]

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

 

[alyafey22]

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 .

 

[alyafey22]

This can be seen by

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

The sum is always 1.

 

[kaliprasad]

This can be seen by

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

The sum is always 1.

I realize

it was my goofup
thanks