Trigonometric Inequalities: Problems with Solutions

Solution:
In each quadrant we look at the unit circle and compare the signs and sizes of the three functions.



Wherever [tex]\tan x[/tex] is defined,
[tex]\sin x\cdot\cos x=\dfrac{\sin x}{\cos x}\cdot\cos^{2}x=\tan x\cdot\cos^{2}x[/tex],
and inside any quadrant [tex]0\lt \cos^{2}x<1[/tex].

Quadrant I: [tex]\tan x>0[/tex]. A positive number multiplied by [tex]\cos^{2}x<1[/tex] becomes smaller, so [tex]\sin x\cdot\cos x\lt \tan x[/tex]. The inequality does not hold.

Quadrant II: [tex]\sin x>0[/tex] and [tex]\cos x<0[/tex], so [tex]\tan x<0[/tex]. A negative number multiplied by [tex]\cos^{2}x<1[/tex] moves closer to 0, i.e. it becomes larger, so [tex]\sin x\cdot\cos x>\tan x[/tex]. The inequality holds.



Quadrant III: [tex]\sin x<0[/tex] and [tex]\cos x<0[/tex], so [tex]\tan x>0[/tex] and, as in Quadrant I, [tex]\sin x\cdot\cos x\lt \tan x[/tex]. The inequality does not hold.

Quadrant IV: [tex]\sin x<0[/tex] and [tex]\cos x>0[/tex], so [tex]\tan x<0[/tex] and, as in Quadrant II, [tex]\sin x\cdot\cos x>\tan x[/tex]. The inequality holds.

Answer: in quadrants II and IV.

Solution:
The fraction is defined only where [tex]\tan x[/tex] exists and is not 0, so [tex]x\neq 0,\ \dfrac{\pi}{2},\ \pi,\ \dfrac{3\pi}{2},\ 2\pi[/tex].

Since [tex]\cos x\le 1[/tex], the numerator is never positive: [tex]\cos x-1\le 0[/tex], and it equals 0 only for [tex]x=0[/tex] and [tex]x=2\pi[/tex], which are already excluded. So on the domain the numerator is negative.

A fraction with a negative numerator is positive exactly when its denominator is negative:
[tex]\dfrac{\cos x-1}{\tan x}>0\Longleftrightarrow \tan x<0[/tex]

The tangent is negative in Quadrant II and in Quadrant IV (see the figure below):
[tex]x\in\left(\dfrac{\pi}{2},\pi\right)[/tex] or [tex]x\in\left(\dfrac{3\pi}{2},2\pi\right)[/tex]

Answer: [tex]x\in\left(\dfrac{\pi}{2},\pi\right)\cup\left(\dfrac{3\pi}{2},2\pi\right)[/tex]

Solution:
[tex]\sin x-\dfrac{1}{2}>0\Longleftrightarrow \sin x>\dfrac{1}{2}[/tex]

We draw the graphs of [tex]y_{1}=\sin x[/tex] and [tex]y_{2}=\dfrac{1}{2}[/tex] (see the figure below).

The graphs intersect where [tex]\sin x=\dfrac{1}{2}[/tex], that is, at [tex]x=\dfrac{\pi}{6}[/tex] and [tex]x=\dfrac{5\pi}{6}[/tex].

Between these two points the sine curve lies above the line [tex]y=\dfrac{1}{2}[/tex]; for [tex]x\in\left[0,\dfrac{\pi}{6}\right]\cup\left[\dfrac{5\pi}{6},2\pi\right][/tex] it does not.

Answer: [tex]x\in\left(\dfrac{\pi}{6},\dfrac{5\pi}{6}\right)[/tex]

Solution:
[tex]\cos x+\dfrac{1}{3}\ge 0\Longleftrightarrow \cos x\ge -\dfrac{1}{3}[/tex]

We draw the graphs of [tex]y_{1}=\cos x[/tex] and [tex]y_{2}=-\dfrac{1}{3}[/tex] on [tex]\left[0,2\pi\right][/tex] (see the figure below).

The graphs intersect where [tex]\cos x=-\dfrac{1}{3}[/tex]. Since [tex]\cos\left(\pi-t\right)=\cos\left(\pi+t\right)=-\cos t[/tex], the solutions of this equation in [tex]\left[0,2\pi\right][/tex] are
[tex]x_{1}=\pi-\arccos\dfrac{1}{3}\approx 1.91[/tex] and [tex]x_{2}=\pi+\arccos\dfrac{1}{3}\approx 4.37[/tex]

The cosine curve is on or above the line [tex]y=-\dfrac{1}{3}[/tex] from [tex]0[/tex] to [tex]x_{1}[/tex] and from [tex]x_{2}[/tex] to [tex]2\pi[/tex].

Answer: [tex]x\in\left[0,\ \pi-\arccos\dfrac{1}{3}\right]\cup\left[\pi+\arccos\dfrac{1}{3},\ 2\pi\right][/tex]

Solution:
[tex]8\left|\tan x\right|-1<0\Longleftrightarrow \left|\tan x\right|\lt \dfrac{1}{8}\Longleftrightarrow -\dfrac{1}{8}\lt \tan x\lt \dfrac{1}{8}[/tex]

We draw the graphs of [tex]y_{1}=\left|\tan x\right|[/tex] and [tex]y_{2}=\dfrac{1}{8}[/tex] (see the figure below).

On the interval [tex]\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)[/tex] the tangent is increasing, and [tex]\tan x=\pm\dfrac{1}{8}[/tex] for [tex]x=\pm\arctan\dfrac{1}{8}[/tex] (the arctangent is an odd function, so [tex]\arctan\left(-\dfrac{1}{8}\right)=-\arctan\dfrac{1}{8}[/tex]).

So on this interval the solution is [tex]x\in\left(-\arctan\dfrac{1}{8},\arctan\dfrac{1}{8}\right)[/tex].

The period of the tangent is [tex]\pi[/tex], so we add [tex]k\pi[/tex].

Answer: [tex]x\in\left(-\arctan\dfrac{1}{8}+k\pi,\ \arctan\dfrac{1}{8}+k\pi\right),\ k\in\mathbb{Z}[/tex]

Solution:
Since [tex]-1\le\sin 2x\le 1[/tex], the denominator satisfies [tex]0\le 1-\sin 2x\le 2[/tex], so it is never negative.

The denominator is 0 when [tex]\sin 2x=1[/tex], i.e. [tex]2x=\dfrac{\pi}{2}+2k\pi[/tex], [tex]x=\dfrac{\pi}{4}+k\pi[/tex]. In [tex]\left(0,2\pi\right][/tex] these are [tex]x=\dfrac{\pi}{4}[/tex] and [tex]x=\dfrac{5\pi}{4}[/tex], which must be excluded.

Everywhere else the denominator is positive, so the fraction is negative exactly when the numerator is negative:
[tex]\cos x<0\Longleftrightarrow x\in\left(\dfrac{\pi}{2},\dfrac{3\pi}{2}\right)[/tex]

From this interval we remove [tex]x=\dfrac{5\pi}{4}[/tex] ([tex]\dfrac{\pi}{4}[/tex] is not in it anyway).

Answer: [tex]x\in\left(\dfrac{\pi}{2},\dfrac{3\pi}{2}\right)\setminus\left\{\dfrac{5\pi}{4}\right\}[/tex], i.e. [tex]x\in\left(\dfrac{\pi}{2},\dfrac{5\pi}{4}\right)\cup\left(\dfrac{5\pi}{4},\dfrac{3\pi}{2}\right)[/tex]

Solution:
We expand the left side with the sine difference formula:
[tex]\sin\left(x-\dfrac{\pi}{3}\right)=\sin x\cos\dfrac{\pi}{3}-\cos x\,\sin\dfrac{\pi}{3}=\dfrac{1}{2}\sin x-\dfrac{\sqrt{3}}{2}\cos x[/tex]

So the inequality becomes
[tex]\dfrac{1}{2}\sin x-\dfrac{\sqrt{3}}{2}\cos x>\sin x[/tex]
Multiplying by 2:
[tex]\sin x-\sqrt{3}\cos x>2\sin x[/tex]
[tex]\sin x+\sqrt{3}\cos x<0[/tex]

Since [tex]\sin x+\sqrt{3}\cos x=2\left(\dfrac{1}{2}\sin x+\dfrac{\sqrt{3}}{2}\cos x\right)=2\sin\left(x+\dfrac{\pi}{3}\right)[/tex], we get
[tex]\sin\left(x+\dfrac{\pi}{3}\right)<0[/tex]

For [tex]x\in\left[0,2\pi\right][/tex] we have [tex]x+\dfrac{\pi}{3}\in\left[\dfrac{\pi}{3},\dfrac{7\pi}{3}\right][/tex], and in this range the sine is negative only for [tex]x+\dfrac{\pi}{3}\in\left(\pi,2\pi\right)[/tex].

Hence [tex]x\in\left(\dfrac{2\pi}{3},\dfrac{5\pi}{3}\right)[/tex].

Answer: [tex]x\in\left(\dfrac{2\pi}{3},\dfrac{5\pi}{3}\right)[/tex]

Solution:
Since the point [tex]\left(p,q\right)[/tex] lies on the circle, [tex]p^{2}+q^{2}=r^{2}[/tex], so there is an angle [tex]\theta[/tex] with
[tex]p=r\cos\theta,\quad q=r\sin\theta[/tex] (see the figure).



Dividing the inequality by [tex]r>0[/tex]:
[tex]\cos\theta\,\sin x-\sin\theta\cos x>\dfrac{1}{2}\Longleftrightarrow \sin\left(x-\theta\right)>\dfrac{1}{2}[/tex]

Therefore
[tex]x-\theta\in\left(\dfrac{\pi}{6}+2n\pi,\ \dfrac{5\pi}{6}+2n\pi\right),\quad n\in\mathbb{Z}[/tex]
[tex]x\in\left(\theta+\dfrac{\pi}{6}+2n\pi,\ \theta+\dfrac{5\pi}{6}+2n\pi\right)[/tex]

Comparing with [tex]x\in\left(\dfrac{\pi}{3},\pi\right)[/tex]: [tex]\theta+\dfrac{\pi}{6}=\dfrac{\pi}{3}[/tex] and [tex]\theta+\dfrac{5\pi}{6}=\pi[/tex], so [tex]\theta=\dfrac{\pi}{6}[/tex] (up to a multiple of [tex]2\pi[/tex]).

Then
[tex]\dfrac{p}{q}=\dfrac{r\cos\theta}{r\sin\theta}=\cot\theta=\cot\dfrac{\pi}{6}=\sqrt{3}[/tex]

Answer: [tex]\dfrac{p}{q}=\sqrt{3}[/tex]

Solution:
On the unit circle, [tex]\sin x[/tex] is the [tex]y[/tex]-coordinate of the point that corresponds to [tex]x[/tex]. We draw the horizontal line [tex]y=\dfrac{1}{2}[/tex]; it cuts the circle at the points [tex]M_{0}[/tex] and [tex]M_{1}[/tex].



These points correspond to the solutions of [tex]\sin x=\dfrac{1}{2}[/tex]:
[tex]x=\dfrac{\pi}{6}+2n\pi[/tex] or [tex]x=\dfrac{5\pi}{6}+2n\pi,\ n\in\mathbb{Z}[/tex]

The inequality holds for the points of the arc [tex]M_{0}BM_{1}[/tex], which lies on or above the line, going counterclockwise from [tex]M_{0}[/tex] to [tex]M_{1}[/tex].

Answer: [tex]x\in\left[\dfrac{\pi}{6}+2n\pi,\ \dfrac{5\pi}{6}+2n\pi\right],\ n\in\mathbb{Z}[/tex]

Solution:
We graph [tex]f(x)=\cos x[/tex] and [tex]g(x)=\dfrac{\sqrt{2}}{2}[/tex] on the interval [tex]\left[0,2\pi\right][/tex].



The graphs intersect where [tex]\cos x=\dfrac{\sqrt{2}}{2}[/tex], that is, at [tex]x=\dfrac{\pi}{4}[/tex] and [tex]x=\dfrac{7\pi}{4}[/tex].

The graph shows that [tex]f(x)\ge g(x)[/tex] (the shaded parts) for
[tex]x\in\left[0,\dfrac{\pi}{4}\right]\cup\left[\dfrac{7\pi}{4},2\pi\right][/tex]

Answer: [tex]x\in\left[0,\dfrac{\pi}{4}\right]\cup\left[\dfrac{7\pi}{4},2\pi\right][/tex]

Solution:
Using [tex]\sin 2x=2\sin x\cos x[/tex] we get
[tex]2\sin x\cos x-6\cos x>0[/tex]
[tex]2\cos x\left(\sin x-3\right)>0[/tex]

Since [tex]\sin x\le 1[/tex], the factor [tex]\sin x-3[/tex] is always negative. So the product is positive exactly when [tex]\cos x<0[/tex].

This agrees with the graphs of [tex]f(x)=\sin 2x[/tex] and [tex]g(x)=6\cos x[/tex] on [tex]\left[0,2\pi\right][/tex]: [tex]f(x)>g(x)[/tex] for [tex]x\in\left(\dfrac{\pi}{2},\dfrac{3\pi}{2}\right)[/tex].



Both functions repeat every [tex]2\pi[/tex], so the solution repeats every [tex]2\pi[/tex] too.

Answer: [tex]x\in\left(\dfrac{\pi}{2}+2n\pi,\ \dfrac{3\pi}{2}+2n\pi\right),\ n\in\mathbb{Z}[/tex]

Solution:
The tangent has period [tex]\pi[/tex], so it is enough to solve the inequality on one period, [tex]\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)[/tex], and then add [tex]n\pi[/tex].

On this interval the tangent is increasing, [tex]\tan\dfrac{\pi}{4}=1[/tex] and [tex]\tan x\to+\infty[/tex] as [tex]x\to\dfrac{\pi}{2}[/tex]. Hence [tex]\tan x\ge 1[/tex] ⇔ [tex]x\in\left[\dfrac{\pi}{4},\dfrac{\pi}{2}\right)[/tex]. The point [tex]\dfrac{\pi}{2}[/tex] is excluded because [tex]\tan\dfrac{\pi}{2}[/tex] is not defined.

On the unit circle (see the figure below) these are the points of the arc from [tex]\dfrac{\pi}{4}[/tex] to [tex]\dfrac{\pi}{2}[/tex] and, one period later, from [tex]\dfrac{5\pi}{4}[/tex] to [tex]\dfrac{3\pi}{2}[/tex].

Answer: [tex]x\in\left[\dfrac{\pi}{4}+n\pi,\ \dfrac{\pi}{2}+n\pi\right),\ n\in\mathbb{Z}[/tex]

Solution:
We look at the graphs of [tex]f(x)=\sin x[/tex] and [tex]g(x)=\cos x[/tex] and first find their intersection points:
[tex]\sin x=\cos x\Longleftrightarrow \tan x=1[/tex] (we may divide by [tex]\cos x[/tex]: if [tex]\cos x=0[/tex], then [tex]\sin x=\pm 1\neq\cos x[/tex]).

In [tex]\left[0,2\pi\right][/tex] this gives [tex]x_{1}=\dfrac{\pi}{4}[/tex], [tex]x_{2}=\dfrac{5\pi}{4}[/tex].



The figure shows that [tex]f(x)>g(x)[/tex] (the shaded region) for [tex]x\in\left(\dfrac{\pi}{4},\dfrac{5\pi}{4}\right)[/tex].

Answer: [tex]x\in\left(\dfrac{\pi}{4},\dfrac{5\pi}{4}\right)[/tex]

Solution:
We graph [tex]f(x)=\sqrt{3}\cos x[/tex] and [tex]g(x)=1+\sin x[/tex] on [tex]\left[0,2\pi\right][/tex] and look for their intersection points [tex]x_{1}[/tex] and [tex]x_{2}[/tex].



[tex]f(x)=g(x)\Longleftrightarrow \sqrt{3}\cos x=1+\sin x\Longleftrightarrow \sin x-\sqrt{3}\cos x=-1[/tex]

Since [tex]\sin x-\sqrt{3}\cos x=2\left(\dfrac{1}{2}\sin x-\dfrac{\sqrt{3}}{2}\cos x\right)=2\sin\left(x-\dfrac{\pi}{3}\right)[/tex],
[tex]\sin\left(x-\dfrac{\pi}{3}\right)=-\dfrac{1}{2}[/tex]

For [tex]x\in\left(0,2\pi\right)[/tex] we have [tex]x-\dfrac{\pi}{3}\in\left(-\dfrac{\pi}{3},\dfrac{5\pi}{3}\right)[/tex], so
[tex]x-\dfrac{\pi}{3}=-\dfrac{\pi}{6}[/tex] or [tex]x-\dfrac{\pi}{3}=\dfrac{7\pi}{6}[/tex]
[tex]x_{1}=\dfrac{\pi}{6},\quad x_{2}=\dfrac{3\pi}{2}[/tex]

The figure shows that [tex]f(x)\lt g(x)[/tex] between [tex]x_{1}[/tex] and [tex]x_{2}[/tex] (check: at [tex]x=\pi[/tex] we get [tex]-\sqrt{3}<1[/tex]).

Answer: [tex]x\in\left(\dfrac{\pi}{6},\dfrac{3\pi}{2}\right)[/tex]