-2
$\begingroup$

No solution, please. Just hint what I should do to get to a simplification.

$$e^{2x}\le e^x+6$$

$$\sin\left(4x\right)<\cos\left(4x\right)$$

$\endgroup$

closed as off-topic by Jyrki Lahtonen, Martin R, Daniele Tampieri, José Carlos Santos, Ernie060 Oct 10 at 12:19

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Jyrki Lahtonen, Martin R, Daniele Tampieri, José Carlos Santos, Ernie060
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ You should include something about what you know of solving such problems, and where you are stuck. This will help answerers avoid wasting time (theirs and yours) telling you things you already know; also, it can help them avoid giving you too much of a hint and spoiling your fun. $\endgroup$ – Blue Oct 9 at 18:29
  • $\begingroup$ Are these a system of inequalities that you are trying to combine, or separate? If they are together use complex exponential representations of sines and cosines. $\endgroup$ – Legorhin Oct 9 at 18:39
  • $\begingroup$ @Durabys You need to mention simultaneous solution' and need to put and' between the two inequations. $\endgroup$ – Dr Zafar Ahmed DSc Oct 10 at 5:53
  • $\begingroup$ They are separate inequalities. $\endgroup$ – Durabys Oct 10 at 17:28
0
$\begingroup$

The first: Consider it a quadratic inequality in $y=e^x$.

The second: Shift one of the arguments so that both are sines or cosines and then apply an appropriate trigonometric formula.

$\endgroup$
  • $\begingroup$ Thanks, man! This what I needed. $\endgroup$ – Durabys Oct 10 at 19:30
0
$\begingroup$

$z=e^x$

The first one becomes a quadratic

In second one divide by cos (4x) which becomes$tan(4x)<0$ but here cos(4x) >0

Find the intersection of this inequality Then take case when cos(4x)<0 then $tan(4x)>0$ find intersection of this also.Then u'll get the inequality

$\endgroup$
  • $\begingroup$ The second hint is half wrong, you do not compare to zero afterwards. What about $\cos 4x<0$? $\endgroup$ – LutzL Oct 10 at 5:42
-1
$\begingroup$

By $t=e^x>0$ we have

$$e^{2x}\le e^x+6\iff t^2-t-6\le0$$

and

$$\sin\left(4x\right)<\cos\left(4x\right)\iff \cos\left(4x\right)\left(\tan\left(4x\right)-1\right)<0$$

$\endgroup$
-1
$\begingroup$

hint

For the first, observe that $$e^{2x}-e^x-6=(e^x+2)(e^x-3)$$

and that $$(\forall x\in \Bbb R) \;\; e^x+2>0$$

For the second. Put $z=4x$. Then, the inequation can be written as

$$\sin(z)-\cos(z)<0$$

or

$$\sqrt{2}\Bigl(\frac{\sin(z)}{\sqrt{2}}-\frac{\cos(z)}{\sqrt{2}}\Bigr)<0$$

which is equivalent to

$$\sin(z-\frac{\pi}{4})<0$$

thus

$$z-\frac{\pi}{4} \in \bigcup_{k\in \Bbb Z}(-\pi+2k\pi,0+2k\pi)$$ and

$$4x\in \bigcup_{k\in \Bbb Z}(-\frac{3\pi}{4}+2k\pi,\frac{\pi}{4}+2k\pi)$$

You can finish.

$\endgroup$
-1
$\begingroup$

The domain for $$f(x)=e^{2x}-e^x-6<0 \implies (e^x-3)(e^2+2)<0$$ is $x <\ln 3.$ $f(-\infty)=-6, f'(x)=e^{x}(2e^x-1)=0 \implies x=-\ln 2, f''(-\ln 2) = 1/2>0, f_{min} =-\frac{25}{4}$. So for $x<-\ln 2$, the range of $f(x)$ is $R_1: (-25/4,-6)$.

Next, the function $g(x)=\sin(4x) -\cos(4x)$ is periodic and bounded in $R_2 :[-\sqrt{2},\sqrt{2}],$ as $R_1 \cap R_2=\phi$ (null), so there will be no real root. See below corves $f(x)$ (red) anf $g(x)$ (blue) they do not intersect for $x<0$, or more precisely for $x<\ln 3$.

enter image description here

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.