Trigonometric Inequalities: Problems with Solutions

Solution:
In each of the quadrants we draw the graphs

In Quadrant I we can see that

$\sin x\cos x>\tan x$ equals
(decimal (+) less than 1) × (decimal (+) less 1) < (decimal (+)),
this way, the product $\sin x \cdot \cos x$ is (+) but it's less than $\tan x$,
which is also (+) due to the positive decimals comparison criterion.

Conclusion: $\sin x \cdot \cos x>\tan x$ is not met.

In Quadrant II we can see that $\sin x\cos x>\tan x$ equals (decimal (+) less than 1) × (decimal (-) less than 1) > (decimal (-)),

so the product $\sin x\cos x$ is (-) but greater than $\tan x$,

which is also (-) due to the negative numbers comparison criterion.

Conclusion: $\sin x\cos x>\tan x$ is met.



Quadrant III we can see that $\sin x\cos x>\tan x$ equals
(decimal (-) less than 0) × (decimal (+) less than 1) < (decimal (+)),
this way, the product $\sin x\cos x$ is (-) and $\tan x$ is (+).

Conclusion: $\sin x\cos x>\tan x$ is not met.

In Quadrant IV we can see that
$\sin x\cos x>\tan x$ equals

(decimal (-) less than 0) × (decimal (+) less than 1) > (decimal (-)),
which means the product $\sin x\cos x$ is (-) but greater than $\tan x$ which is also (-) due to the negative numbers comparison criterion.

Conclusion: $\sin x\cos x>\tan x$ is met. The answer is:
The inequality $\sin x\cos x>\tan x$ is met in quadrants II and IV.

Solution:
In order to solve this inequality we have to analyze the behavior of the expressions in each quadrant of the unit circle.

$\frac{\cos x-1}{\tan x}>0 \Longrightarrow \cos x-1<0$ and $\tan x$

Since $\cos x-1<0$ if $\cos x<0$

This only happens in Quadrant II i.e. when $x\in \left( \frac{\pi }{2},\pi \right)$

We can see that $\cos x-1<0$ which means $\cos x$ is (+) but

$\cos x<1$ (by the definition of the cosene) and that $\tan x$ is (-)

in Quadrant IV, i.e. when $x\in \left( \frac{3\pi }{2},2\pi \right) $
Finally, the solution set will be: $x\in \left( \frac{\pi }{2},\pi \right) \cup \left( \frac{3\pi }{2},2\pi \right) $

Solution:
$\sin x-\frac{1}{2}>0\Longrightarrow \sin x>\frac{1}{2}$ we draw the graph with $y_{1}=\sin x\qquad y_{2}=\frac{1}{2}$



The intersection occurs when $x=\frac{\pi }{6}$ and $x=\frac{5\pi }{6}$

So the possible solution to the given inequality is $\frac{\pi }{6} < x < \frac{5\pi }{6}$

Solution:
First, we draw the graph $y_{1}=\cos x\qquad y_{2}=-\frac{1}{3}$



To intersect $y_{1}$ and $y_{2}$ we have that $\cos x=-\frac{1}{3}\Longrightarrow x=\arccos \left( -\frac{1}{3}\right) $

and since cosine is an even function, we have that $x=\pm \arccos \left(\frac{1}{3}\right) $

This way, the interval representing the main solution of the inequality

$\cos x+\frac{1}{3}\geq 0$ is $-\arccos \left( \frac{1}{3}\right) \leq x\leq \arccos \left( \frac{1}{3}\right) $ when $0\leq x\leq 2\pi $

Solution:
Because $8\left\vert \tan x\right\vert -1<0\Longrightarrow \left\vert \tan x\right\vert <\frac{1}{8}$ and by the definition of the absolute value we have that

$-\frac{1}{8}<\tan x<\frac{1}{8}$ now, let us see the functions $y_{1}=\left\vert \tan x\right\vert \ \ y_{2}=\frac{1}{8}$ in the graph



Since $\left\vert \tan x\right\vert =\frac{1}{8}\Longrightarrow x=\pm \arctan \frac{1}{8}$, the solution set for the given inequalities will be

$\arctan \left( -\frac{1}{8}\right) < x < \arctan \frac{1}{8}$ and provided that the $\arctan x$ function is odd we have that

$-\arctan \left( \frac{1}{8}\right) < x < \arctan \frac{ 1}{8}$ and since the period of the tangent function is $\pi $,

then the generalized period will be $k\pi $ where $k=0,\pm 1,\pm 2,\pm 3,....$ Finally, the generalized solution is:

$-\arctan \left( \frac{1}{8}\right) +k\pi < x < \arctan \frac{1}{8}+k\pi \qquad k=0,\pm 1,\pm 2,\pm 3,....$

Solution:
We know that $-1\leq \sin 2x\leq 1\Longrightarrow 1\geq -\sin 2x\geq -1$

so, $2\geq 1-\sin 2x\geq 0$ then, we can say that $\frac{\cos x}{1-\sin 2x} < 0\Longrightarrow $ $\cos x<0$

and that $1-\sin 2x\neq 0\Longrightarrow x\in \left( \frac{\pi }{2},\frac{3\pi }{2}\right) $ and also that $1\neq \sin 2x\Longrightarrow 2x\neq \left\{ \frac{\pi }{2},\frac{5\pi }{2},...\right\} $.

Finally, $x\neq \left\{ \frac{\pi }{4},\frac{5\pi }{4},...\right\} \Longrightarrow x\in \left( \frac{\pi }{2},\frac{3\pi }{2}\right) -\left\{ \frac{5\pi }{4}\right\} $

Solution:
$\sin \left( x-\frac{\pi }{3}\right) >\sin x\Longrightarrow \left[ \sin x\cos \frac{\pi }{3}-\sin \frac{\pi }{3}\cos x\right] >\sin x$

Then, $\frac{1}{2}\sin x-\frac{\sqrt{3}}{2}\cos x>\sin x\Longrightarrow \sin x-\sqrt{3}\cos x>2\sin x$

So, $0>\sin x+\sqrt{3}\cos x\Longrightarrow 0>2\sin \left( x+\frac{\pi }{3}\right) \Longrightarrow 2\sin \left( x+\frac{\pi }{3}\right) < 0$

And finally, we have that $\left( x+\frac{\pi }{3}\right) \in \left( \pi ,2\pi \right) \Longrightarrow x\in \left( \frac{2\pi }{3},\frac{5\pi }{3}\right) $

Solution:
$p\sin x-q\cos x>\frac{r}{2}$ $\Longrightarrow \frac{p}{r}\sin x-\frac{q}{r}\cos x>\frac{1}{2}$

and provided that $\left( p,q\right) $ is a point that belongs to the circumference

then, $\frac{p}{r}=\cos \theta $ and $\frac{q}{r}=\sin \theta $ so, $\cos \theta \sin x-\sin \theta \cos x>\frac{1}{2}$

This tells us that $\sin \left( x-\theta \right) >\frac{1}{2}$

then, $2n\pi +\frac{\pi }{6} < x-\theta <\frac{5\pi }{6}+2n\pi$ with $n\in Z$

given $\frac{\pi }{3} < x < \pi$.
Finally, $\theta -\frac{\pi }{6}-2n\pi \Longrightarrow \cot \theta -\frac{p}{q}-\cot \frac{\pi }{6}-\sqrt{3}$

so, $\frac{p}{q}=\sqrt{3}$

Solution:
By using the unit circle, we can locate the points
$M_{0}$ and $M_{1}$, if we take the solution $\sin x=\frac{1}{2}$ then we have that

$x=\frac{\pi }{6}+2n\pi $ or $x=\frac{5\pi }{6}+2n\pi $ where $n\in Z$



So, the solution to the given inequality are all values of $x$ in the interval $\frac{\pi }{6}+2n\pi \leq x\leq \frac{5\pi }{6}+2n\pi$

Solution:
If we graph the functions $f(x)=\cos x\qquad g(x)=\frac{\sqrt{2}}{2}$
in the interval $0\leq x\leq 2\pi $



And by intersecting the functions $f$ and $g$ we have that $\cos x=\frac{\sqrt{2}}{2}$

this happens if $x=\frac{\pi }{4}$ or $x=\frac{7\pi }{4}$

We can see in the graph that $f(x)\geq g(x)$ when

$x\in \left[ 0,\frac{\pi }{4}\right] \cup \left[ \frac{7\pi }{4},2\pi \right]$

Solution:
First, we define the functions $f(x)=\sin 2x\qquad g(x)=6\cos x$

and we draw the graph in the interval $\left[ 0,2\pi \right] $




Then, we can see in the graph that $f(x)>g(x)$ in the interval

$\frac{\pi }{2} < x < \frac{3\pi }{2}$ and by generalizing in any of the periods of the functions

we have that $x\in \left( 2n\pi +\frac{\pi }{2};2n\pi +\frac{3\pi }{2}\right) $ where $n\in Z$

Solution:
By drawing the graph of the equation in a circumference, we remember that the domain of the tangent is $\left( -\frac{\pi }{2};\frac{\pi }{2}\right) \cup \left( \frac{\pi }{2};\frac{3\pi }{2}\right) $



The tangent is greater than $1$ when $x$ is in the interval $\frac{\pi }{4} < x < \frac{\pi }{2}$ and also in the interval $\frac{5\pi }{4} < x < \frac{5\pi }{2}$

and by generalizing in any period we have that $\tan x\geq 1\Leftrightarrow x\in \left[ \frac{\pi }{4}+n\pi ;\frac{5\pi }{2}+n\pi \right] $

Solution:
By looking at the graphs of the functions $f(x)=\sin x$ and $g(x)=\cos x$

we find the intersection points $\sin x$ $=\cos x\Longrightarrow \tan x=1$

$x=\arctan 1$ then, $x_{1}=\frac{\pi }{4}\qquad x_{2}=\frac{5\pi }{4}$ and if we draw the graph in $\left[ 0;2\pi \right] $



We can see in the image that $f(x)>g(x)$ if $\frac{\pi }{4} < x < \frac{5\pi }{4}$

Solution:
First, we make $f(x)=\sqrt{3}\cos x\qquad g(x)=1+\sin x$ then, we draw the graph in the interval $\left[ 0,2\pi \right]$



And we get the points $x_{1}=x _{2}$ which are the points of intersection for the two functions.

Then, $f(x)=g(x)\Longrightarrow \sqrt{3}\cos x=1+\sin x$

So, $-1=\sin x-\sqrt{3}\cos x=2\sin \left( x-\frac{\pi }{3}\right) \Longrightarrow \sin \left( x-\frac{\pi }{3}\right) =-\frac{1}{2}$

and applying the inverse function we have that $x-\frac{\pi }{3}=\arcsin (-\frac{1}{2})$

so, $x-\frac{\pi }{3}=-\frac{\pi }{6}\qquad $and $x-\frac{\pi }{3}=\frac{7\pi }{6}$

After that, finding $x$ we have that $x_{1}=\frac{\pi }{6}\qquad x_{2}=\frac{3\pi }{2}$

Finally, the solution is $x\in \left( \frac{\pi }{6};\frac{3\pi }{2}\right) $