# Solving trigonometric inequality for positive values

Let us consider equation:

$u=-q\cos(wx)$ where $q$ and $w$ are constants.

I'm looking for values of $x$ where $u$ is positive so I write

$-q\cos(wx)>0.$

So when I solve the sign changes (right?), because of the $-q$, and I get

$\cos(wx)<0.$

Then I take the inverse and end up with

$wx < \pi/2$ as a possible answer (right?),

so then $x < (\pi/2)(1/w),$

then if I assign $x_1=d(\pi/2)(1/x)$, where $0<d<1$, then $x_1$ can be an answer (right?) because $x_1$ will always be less than $(\pi/2)(1/w)$ as required by the inequality.

However, when I plug in $x_1$ into the original $u=-q\cos(wx)$ I end up with

$u=-q\cos(d(\pi/2))$ with $0<d<1,$

but this isn't correct because cosine is positive for angles less than $\pi/2$ but greater than 0, which is what I end up with in the argument, $d(\pi/2)$, and multiplying it by the negative $q$ will result in a negative $u$, which is not what I'm looking for because I'm looking for positive $u$'s.

So there it is, I think im solving the inequality wrong and that's why I'm getting the wrong answer. Please help, I can't solve this inequality.

the solution of the inequation

$cos(wx)<0$ is

$\frac{\pi}{2}+2k\pi<wx<\frac{3\pi}{2}+2k\pi$

which gives

$\frac{1}{w}( \frac{\pi}{2}+2k\pi )<x<\frac{1}{w}( \frac{3\pi}{2}+2k\pi )$.

To solve trigonometric inequations, Always use the trigonometric circle.

• so the answer comes from the fact that cos is negative in quadrant 2 and 3 correct? and thats why the pi/2 and 3pi/2. and the 2kpi is just any full circle correct? so a possible answer to x is (pi/2w)b if k=0 and b=2,6,10,... ? but thats just by looking at the trig circle an seeing where cos ends up neg so itll multiply by the neg q and ill get the pos u. but s there a way to get actual possible numbers by doing math? not just by inspectoin. – uno Oct 16 '16 at 21:16
• yes, i think you see now – hamam_Abdallah Oct 16 '16 at 21:19