MHB What is the Solution to $4\sin(x)+3\cos(x)\geq0$ in Terms of Domain and Range?

[bahadeen]

I need explanation
View attachment 4231

 

[Ackbach]

You're certainly fine up to and including the $4 \sin(x)\ge -3 \cos(x)$ step. And you'd even be quite justified in doing
$$\sin(x)\ge -\frac34 \, \cos(x).$$
The problem is when you divide through by $\cos(x)$. You cannot, in general, keep the inequality the same direction because $\cos(x)$ is not always positive. Whenever it's negative, you should technically reverse the inequality.

However, since you don't know in advance when to do that, I suggest a different approach: find out where $x=\arctan(-3/4)$. That is, find ALL solutions of this equation. Then divide up the real line into pieces depending on where you get the equalities. Finally, sample the expression $4\sin(x)+3 \cos(x)$ inside each of those pieces - find out where it's positive and where it's negative. The positive portions are in the domain, and the negative are not.

Does that make sense?

 

[johng1]

You want to solve $4\sin(x)+3\cos(x)\geq0$. This is the same as $5({4\over5}\sin(x)+{3\over 5}\cos(x))\geq0$. Now there is $\theta$ with $\cos(\theta)={4\over 5}$ and $\sin(\theta)={3\over5}$. Namely $\theta=\arctan(3/4)$. So the original inequality is the same as $5\sin(x+\theta)\geq0$. Now $\sin(x)\geq0$ precisely when $x\in[2k\pi,(2k+1)\pi]$ for an integer $k$. So your inequality is true for $x\in[2k\pi-\theta,(2k+1)\pi-\theta]$.

You might want to use your favorite graphing software to graph the functions to check the above.