Solve $6\sin x \cos 2x\ge 0$

Question

How can I solve the following inequality: $${6\sin x\cos 2x\ge 0}$$ Can you give me an explicit explanation of how this exercise can be understood. I have no problems with trigonometric equations, but when it comes to inequailities of this type, where I have to combine the solutions from two or more inequalities solved apart, I get really confused. Thank you!

Answer 1

If $\sin x\cos2x=0$

either $\sin x=0\implies x=n\pi$ or $\cos2x=0\implies2x$ must be odd multiple of $\dfrac\pi2$

Else we need $(1)\displaystyle\sin x>0$ and $\displaystyle\cos2x>0$ OR $(2)\displaystyle\sin x<0$ and $\displaystyle\cos2x<0$

For $\displaystyle(2), \pi<x<2\pi$ and $\displaystyle2m\pi+\frac\pi2<2x<2m\pi+\frac{3\pi}2\iff m\pi+\frac\pi4<x<m\pi+\frac{3\pi}4$

So, the intersection is $\displaystyle\pi+\frac\pi4<x<\pi+\frac{3\pi}4$

For $\displaystyle(1),0<x<\pi$

and $\left(2m\pi+0<2x<2m\pi+\dfrac\pi2\text{ OR }2m\pi+\dfrac{3\pi}2<x\le2m\pi+2\pi\right)$

$\displaystyle\implies0<x<\pi\text{ and }\left(m\pi<x<m\pi+\dfrac\pi4\text{ OR }m\pi+\dfrac{3\pi}4<x\le m\pi+\pi\right)$

Can you find the intersections?

Answer 2

Hint :

$\cos 2x=f(\sin x)$?? what is that $f$? (you can actually neglect that $6$...)

Answer 3

Hints:

$6\sin x\cos2x\geq0\iff\sin x\cos2x\geq0$ and $\cos2x=1-2\sin^{2}x$

So to be solved is $y\left(1-2y^{2}\right)\geq0$ where $y=\sin x$