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!
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1for that inequality to hold, either both must be positive or both must be negatives, or one of them zero. you need to find the values of $x $ where that condition is satisfied. – S L May 03 '14 at 10:16
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$g(x)=sin(x)cos(2x)$ is an odd and $2\pi$-perodic function so we can solve for $x$ in the interval $[0,\pi]$ and then conclude. – Walid Taamallah May 03 '14 at 10:20
3 Answers
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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?
lab bhattacharjee
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Yes, but does not $cos2x$ have a different period($\pi$) from $sinx$($2\pi$)?How is it possible to combine two intervals with different periods? – wonderingdev May 03 '14 at 10:22
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I have one more question, because , excuse me , but I have some misunderstandings in your explanation. When $cos2x>0$ can we directly write that ${-\pi\over 2}+2\pi k\lt2x\lt{{\pi\over 2} +2\pi k}$. But this means that the period is different from that of $sinx>0$!!?? – wonderingdev May 03 '14 at 10:37
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So, then , how do I find the intersection of two intervals with different periods? – wonderingdev May 03 '14 at 10:42
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1@JohnG., Probably this (http://math.stackexchange.com/questions/164221/period-of-the-sum-product-of-two-functions) will help – lab bhattacharjee May 03 '14 at 10:55
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Hint :
$\cos 2x=f(\sin x)$?? what is that $f$? (you can actually neglect that $6$...)
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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 $y=\sin x$
drhab
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