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Solve this trigonometric inequality: $$ \sin x \sin 2x - \cos x \cos 2x > \sin 6x $$ My steps: $$ \cos x \cos 2x - \sin x \sin 2x < - \sin 6x $$ $$ \cos 3x < \sin (-6x)$$ $$ \cos 3x < \cos (\frac{\pi}{2}+6x) $$ From this we get: $$ 3x > \dfrac{\pi}{2}+6x+2k\pi$$ $$ -3x > \dfrac{\pi}{2}+2k\pi$$ $$ x < -\dfrac{\pi}{6}+ \dfrac{2k\pi}{3}$$

and

$$ 3x < -\dfrac{\pi}{2} - 6x + 2k\pi$$ $$ 9x < -\dfrac{\pi}{2} + 2k\pi$$ $$ x < - \dfrac{\pi}{18} + \dfrac{2k\pi}{9} $$

as you can see the solution is not correct

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$$\cos { \left( 3x \right) < } -\sin { \left( 6x \right) } \\ \cos { \left( 3x \right) < } -2\sin { \left( 3x \right) \cos { \left( 3x \right) } } \\ \cos { \left( 3x \right) } \left( 1+2\sin { \left( 3x \right) } \right) <0$$

$$1.\cos { \left( 3x \right) } >0\quad and\quad 1+2\sin { \left( 3x \right) } <0\\ 2.\cos { \left( 3x \right) } <0\quad and\quad 1+2\sin { \left( 3x \right) } >0\\ $$

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Your error is that in general $\cos A < \cos B$ does not imply that $A > B$. (If you don't see this right away, try $A = \pi$ and $B = \dfrac{3\pi}{2}$).

Instead, try the following approach:

$ \cos x \cos 2x - \sin x \sin 2x < -\sin 6x $

$\cos 3x < -\sin 6x$

$\cos 3x + \sin 6x < 0$

$\cos 3x + 2\sin 3x\cos 3x < 0$

$\cos 3x(1+2\sin 3x) < 0$

So we need exactly one of $\cos 3x$ and $1+2\sin 3x$ to be negative.

We know that $\cos 3x < 0$ when $\dfrac{\pi}{2}+2\pi k < 3x < \dfrac{3\pi}{2}+2\pi k$ for some integer $k$. Also, $1+2\sin 3x < 0$ when $\sin 3x < -\dfrac{1}{2}$, i.e. $\dfrac{7\pi}{6}+2\pi k < 3x < \dfrac{11\pi}{6}+2\pi k$ for some integer $k$.

Can you finish the problem from here?

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  • $\begingroup$ Yes I did, thank you! $\endgroup$ – Gjekaks Aug 10 '15 at 21:14
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$$ \sin (x) \sin (2x) - \cos (x) \cos (2x) > \sin (6x) \Longleftrightarrow$$ $$ -\cos(3x) > \sin (6x) $$

So we find $4$ solutions:

$$x=\frac{2\pi n +\pi}{3}$$ $$\frac{2\pi n -\pi}{3}<x<\frac{12\pi n -5\pi}{18}$$ $$\frac{4\pi n -\pi}{6}<x<\frac{12\pi n -\pi}{18}$$ $$\frac{4\pi n +\pi}{6}<x<\frac{2\pi n +\pi}{3}$$

With $n \in \mathbb{Z}$

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