Another trigonometry inequation?

[Johulus]

Inequation:
\(\displaystyle \sin(x)\, \cdot\, \cos(5x)\, >\, \sin(9x)\, \cdot\, \cos(3x) \)
Using formulas:
\(\displaystyle \sin(x)\cos(y)\, =\dfrac{1}{2}[\sin(x+y)\, +\, \sin(x-y)]\)
And
\(\displaystyle \sin(x)\, +\, \sin(y)\, =\, 2\sin(\dfrac{x+y}{2})\cos(\dfrac{x-y}{2})\)

\(\displaystyle \sin(6x)\, -\, sin(4x)\, >\, \sin(12x)\, +\, \sin(6x) \\ \sin(12x)\, +\, \sin(4x)\, <\, 0 \\ \sin(8x)\cos(4x)\, <\, 0 \)
Going to 2 cases:
\(\displaystyle \sin(8x)\, <\, 0 \, \, \, \sin(8x)\, >\, 0 \\ \cos(4x)\, >\, 0 \, \, \, \cos(4x)\, <\, 0 \)

Can I now determine where sine and cosine for the first case are equal to 0 then draw all points on a circle and then write next to those intervals where function is less and where higher than 0 for sine and cosine using + and - signs and then just determine where the total is less than zero by looking at those intervals and those + and - signs ? If you know what I mean? And just repeat for second case. Or, is there any easier or 'more correct' way?
Because we haven't done any task of this kind.

 

[Ishuda]

If I'm understanding you correctly, yes you can do that. Then after you are finished with that part, realize that sine and cosine repeat every two pi (each time around the circle) and that will give you a set of intervals [but you need to sync the 4x and 8x cycles].

However, there is another way which may prove beneficial in other problems. Going through the first case,
\(\displaystyle \sin(8x)\, <\, 0 \, \, and\, \cos(4x)\, >\, 0\)
we have
sin(8x) is less than zero when 8x is in any open \(\displaystyle \pi\) interval such that
\(\displaystyle \pi + 2 n \pi < 8x < 2 \pi + 2 n \pi\)
or
\(\displaystyle \frac{\pi}{8} + \frac{n \pi}{4} < x < \frac{\pi}{4} + \frac{n \pi}{4}\)

cos(4x) is greater than zero when 4x is in any open \(\displaystyle \pi\) interval such that
\(\displaystyle -\frac{\pi}{2} + 2 m \pi < 4x < \frac{\pi}{2} + 2 m \pi\)
or
\(\displaystyle -\frac{\pi}{8} + \frac{2 m \pi}{4} < x < \frac{\pi}{8} + \frac{2 m \pi}{4}\)

Now, we could start anywhere for the intervals as long as the one with the n's were some \(\displaystyle \frac{\pi}{8} + \frac{j \pi}{4}\) which means the other side of that inequality would be \(\displaystyle \frac{\pi}{4} + \frac{j \pi}{4}\) and the one for the m's were some \(\displaystyle -\frac{\pi}{8} + \frac{2 k \pi}{4}\) with the corresponding values on the other end of the inequalities. If we choose j=k=1 and put everything over a common denominator we have
sin(8x)<0 if
\(\displaystyle \frac{3\pi}{8} + \frac{2 n \pi}{8} < x < \frac{4\pi}{8} + \frac{2 n \pi}{8}\)
and cos(4x)>0 if
\(\displaystyle \frac{3\pi}{8} + \frac{4 m \pi}{8} < x < \frac{5\pi}{8} + \frac{4 m \pi}{8}\)

From those two equations we see that we have to choose the smaller interval for the n to satisfy the sine part and the larger cycle for the m to satisfy the cosine part or
sin(8x) cos(4x) < 0 with sin(8x) <0 on
\(\displaystyle \frac{3\pi}{8} + \frac{4 n \pi}{8} < x < \frac{4\pi}{8} + \frac{4 n \pi}{8}; n\, \epsilon\, \mathcal{Z}\);