Your cart is empty
Trigonometric equations, inequalities
The equations containing a trigonometric function applied to an unknown variable are called trigonometric equations. Their solutions can be given with the help of the definitions and the theorems verified earlier.
Example 1
Let us solve the equation \latex{\sin ^{2}x=\frac{1}{2}} on the set of real numbers.
Solution
Since both sides are non-negative by taking the square root of both sides:
\latex{\sqrt{\sin ^{2}x }=\sqrt{\frac{1}{2} },}
\latex{|\sin x|=\frac{\sqrt{2} }{2}.}
\latex{|\sin x|=\frac{\sqrt{2} }{2}.}
Thus
\latex{\sin x=\frac{\sqrt{2} }{2}} or \latex{\sin x=-\frac{\sqrt{2} }{2}.}
Let us draw a figure in both cases. (Figure 39)
\latex{x_{2}}
\latex{\frac{\sqrt{2} }{2}}
\latex{x_{1}}
\latex{x_{4}}
\latex{-\frac{\sqrt{2} }{2}}
\latex{x_{3}}
\latex{ y }
\latex{ x }
\latex{ x }
\latex{ y }
Figure 39
Based on these the solutions can be seen:
\latex{x_{1}=\frac{\Pi }{4} +k\times 2\Pi ,}
\latex{x_{2}=\frac{3\Pi }{4}+l\times 2\Pi ,}
\latex{x_{3}=-\frac{\Pi }{4}+m\times 2\Pi ,}
\latex{x_{4}=-\frac{3\Pi }{4}+n\times 2\Pi ,}
where k, l, m and n are whole numbers.
By combining the two figures the solutions can also be combined (Figure 40):
By combining the two figures the solutions can also be combined (Figure 40):
\latex{x=\frac{\Pi }{4}+k\times \frac{\Pi }{2},} where \latex{k\in\Z.}
By substitution and checking we can make sure that these are valid roots.
\latex{ y }
\latex{ x }
Figure 40
Example 2
Let us solve the following equation on the set of real numbers:
\latex{\cos x=\sin (x+\frac{2\Pi }{3} ).}
\latex{ x }
\latex{ x }
\latex{ y }
Figure 41
Solution I
Let us use the relation \latex{\cos x=\sin (x+\frac{\Pi }{2} )} (Figure 41), based on whichthe equation can be rewritten as follows:
\latex{\sin (x+\frac{\Pi }{2} )=\sin (x+\frac{2\Pi }{3} ).}
Based on the definition it can easily be seen that the equation \latex{\sin \alpha =\sin\beta} holds if and only if (Figure 42, Figure 43)
\latex{\beta =\alpha + k\times 2\Pi} or \latex{\beta =\Pi -\alpha +k\times 2\Pi (k\in \Z).}
So on one hand from the equation:
\latex{x+\frac{\Pi }{2}=x+\frac{2\Pi }{3} +k\times 2\Pi ,}
is obtained, which does not have a solution.
On the other hand:
On the other hand:
\latex{x+\frac{\Pi }{2}=\Pi -x-\frac{2\Pi }{3}+k\times 2\Pi ,}
\latex{2x=-\frac{\Pi }{6}+k\times 2\Pi ,}
\latex{x=-\frac{\Pi }{12}+k\times \Pi .}
\latex{2x=-\frac{\Pi }{6}+k\times 2\Pi ,}
\latex{x=-\frac{\Pi }{12}+k\times \Pi .}
It is indeed a solution of the equation.
\latex{ x }
\latex{ y }
Figure 42
Solution II
Let us rewrite the right-hand-side of the equation by applying the addition theorem:
\latex{\sin (x+\frac{2\Pi }{3} )=\sin x\times \cos \frac{2\Pi }{3}+\cos x\times \sin \frac{2\Pi }{3}=-\frac{1}{2}\times \sin x+\frac{\sqrt{3} }{2}\times \cos x.}
Thus
\latex{\cos x=-\frac{1}{2}\times \sin x+\frac{\sqrt{3} }{2}\times \cos x,}
\latex{2\times \cos x=-\sin x+\sqrt{3}\times \cos x,}
\latex{\sin x=(\sqrt{3}-2 )\times \cos x.}
\latex{2\times \cos x=-\sin x+\sqrt{3}\times \cos x,}
\latex{\sin x=(\sqrt{3}-2 )\times \cos x.}
Since there is no \latex{ x } for which sin \latex{ x } and cos \latex{ x } would be \latex{ 0 } simultaneously, this equation cannot have a root for which \latex{\cos x=0}. Therefore we can divide both sides of the equation by \latex{\cos x}:
\latex{\tan x=\sqrt{3}-2.} From here \latex{x=-\frac{\Pi }{12}+k\times \Pi ,}
where \latex{ k } is an arbitrary whole number.
\latex{ x }
\latex{ y }
Figure 43
Example 3
Let us solve the following equation on the set of real numbers:
\latex{4\times \sin 2x=3\times \tan x-3\times \cot x.}
Solution
The equation has a meaning only where
\latex{\cos x\neq 0,} i.e. \latex{x\neq \frac{\Pi }{2}+k\times \Pi (k\in \Z);}
\latex{\sin x\neq 0,} i.e. \latex{x\neq l\times \Pi (l\in \Z).}
\latex{\sin x\neq 0,} i.e. \latex{x\neq l\times \Pi (l\in \Z).}
Based on the relation relating to \latex{\sin 2x} and the definition of \latex{\tan x} and \latex{\cot x}:
\latex{4\times 2\times \sin x\times \cos x=\frac{3\times \sin x}{\cos x} -\frac{3\times \cos x}{\sin x}.}
After removing the fractions:
\latex{8\times \sin ^{2}x\times \cos ^{2}x=3\times \sin ^{2}x-3\times \cos ^{2}x.}
Since \latex{\sin ^{2}x=1-\cos ^{2}x,}
\latex{8\times (1-\cos ^{2}x )\times \cos ^{2}x=3\times (1-\cos ^{2}x )-3\times \cos ^{2}x.}
After rearranging:
\latex{8\times \cos ^{4}x-14\times \cos ^{2}x+3=0.}
The equation is quadratic in terms of \latex{\cos ^{2}x;} its solutions are:
\latex{\cos ^{2}x=\frac{3}{2}} or \latex{\cos ^{2}x=\frac{1}{4}.}
No solution is obtained in the first case, since \latex{\cos ^{2}x\leq 1.}
In the second case \latex{|\cos x|=\frac{1}{2}} i.e.
In the second case \latex{|\cos x|=\frac{1}{2}} i.e.
\latex{\cos x=\frac{1}{2}} or \latex{\cos x=-\frac{1}{2},}
\latex{x_{1}=\pm \frac{\Pi }{3}+2k\Pi (k\in \Z), x_{2}=\pm \frac{2\Pi }{3}+2l\Pi (l\in \Z).}
\latex{x_{1}=\pm \frac{\Pi }{3}+2k\Pi (k\in \Z), x_{2}=\pm \frac{2\Pi }{3}+2l\Pi (l\in \Z).}
Both systems of roots (Figure 44, Figure 45) satisfy the original equation.
\latex{ x }
\latex{ y }
Figure 44
Example 4
Let us find the roots of the following equation: \latex{\sin x+\sqrt{3}\times \cos x=1.}
\latex{ x }
\latex{ y }
Figure 45
Solution I
It is an effective method for such equations if we try to get the sum of squares of the coefficients of the two terms on the left-hand-side to be equal to \latex{ 1 }. For this reason let us divide both sides of the equation by
\latex{\sqrt{1^{2}+(\sqrt{3} )^{2} }=2:}
\latex{\frac{1}{2}\times \sin x+\frac{\sqrt{3} }{2}\times \cos x=\frac{1}{2}.}
In this case the coefficients can be considered as the sine and the cosine of a suitably selected angle:
\latex{\sin \varphi= \frac{1}{2},} at the same time \latex{\cos \varphi=\frac{\sqrt{3} }{2}.}
This angle is unambiguously defined in the unit circle: \latex{\varphi=\frac{\Pi }{6}} (Figure 46). So we can rewrite the equation to the following form:
\latex{\sin \frac{\Pi }{6}\times \sin x+\cos \frac{\Pi }{6}\times \cos x=\frac{1}{2}.}
The expression on the left-hand-side can be rewritten by using the addition theorem:
\latex{\cos (x-\frac{\Pi }{6} )=\frac{1}{2},}
\latex{x-\frac{\Pi }{6}=\pm \frac{\Pi }{3}+k\times 2\Pi (k\in \Z).}
\latex{x_{1}=\frac{\Pi }{2}+2k\Pi} and \latex{x_{2}=-\frac{\Pi }{6}+2l\Pi (k,l\in \Z).}
\latex{x_{1}=\frac{\Pi }{2}+2k\Pi} and \latex{x_{2}=-\frac{\Pi }{6}+2l\Pi (k,l\in \Z).}
As during the transformations equivalent equations are obtained, the results obtained will indeed be solutions.
The equation of the example can also be solved by squaring both sides. However in this case it is very important to check all results obtained, since this step is not an equivalent transformation. As in this case the set of roots might be extended and false roots might arise.
The equation of the example can also be solved by squaring both sides. However in this case it is very important to check all results obtained, since this step is not an equivalent transformation. As in this case the set of roots might be extended and false roots might arise.
The equation \latex{x=2} has only one solution, but the equation \latex{x^{2}=4} obtained after the squaring has already two solutions: \latex{x_{1,2}=\pm 2}
\latex{ x }
\latex{ y }
Figure 46
Solution II
By squaring both sides of the equation:
\latex{\sin ^{2}x+3\times \cos ^{2}x+2\times \sqrt{3}\times \sin xx\times \cos \times =1.}
Let us substitute \latex{\sin ^{2}x+\cos ^{2}x} for the \latex{ 1 } on the right-hand-side of the equation.
\latex{\sin ^{2}x+3\times \cos ^{2}x+2\times \sqrt{3}\times \sin x\times \cos x=\sin ^{2}x+\cos ^{2}x,}
\latex{2\times \cos ^{2}x+2\times \sqrt{3}\times \sin x\times \cos x=0,}
\latex{\cos x\times (\cos x+\sqrt{3}\times \sin x )=0.}
\latex{2\times \cos ^{2}x+2\times \sqrt{3}\times \sin x\times \cos x=0,}
\latex{\cos x\times (\cos x+\sqrt{3}\times \sin x )=0.}
Two possibilities are obtained from here:
\latex{\cos x=0} and \latex{\cos x+\sqrt{3}\times \sin x=0.}
The following is obtained from the first case:
\latex{x=\frac{\Pi }{2}+k\times 2\Pi (k\in \Z).}
however when checking only
\latex{x_{1}=\frac{\Pi }{2}+k\times 2\Pi (k\in \Z).}
will be a solution.
Based on the other equation \latex{\tan x=-\frac{\sqrt{3} }{3}} is obtained, which implies \latex{x=-\frac{\Pi }{6}+l\Pi (l\in \Z).} Out of these only \latex{x_{2}=-\frac{\Pi }{6}+l\times 2\Pi} will be a solution of the equation, which we can make sure of by checking.
Based on the other equation \latex{\tan x=-\frac{\sqrt{3} }{3}} is obtained, which implies \latex{x=-\frac{\Pi }{6}+l\Pi (l\in \Z).} Out of these only \latex{x_{2}=-\frac{\Pi }{6}+l\times 2\Pi} will be a solution of the equation, which we can make sure of by checking.
Example 5
For which real numbers is the following inequality true: \latex{\sin 2x\leq \frac{1}{2}} ?
Solution II
Let us represent the possible values for \latex{2x} in the unit circle.
In Figure 47 it can be seen that the condition \latex{\sin 2x=\frac{1}{2}} is satisfied inthe case of
In Figure 47 it can be seen that the condition \latex{\sin 2x=\frac{1}{2}} is satisfied inthe case of
\latex{2x=\frac{\Pi }{6}} and \latex{2x=\frac{5\Pi }{6}.}
By giving the conditions satisfying the inequality we have to pay attention to correctly select the end-points of the given intervals.
I.e. the limit on the left can be \latex{\frac{5\Pi }{6},} but instead of \latex{\frac{\Pi }{6}} we should consider \latex{\frac{\Pi }{6}+2\Pi =\frac{13\Pi }{6}} as the limit on the right. I.e.:
I.e. the limit on the left can be \latex{\frac{5\Pi }{6},} but instead of \latex{\frac{\Pi }{6}} we should consider \latex{\frac{\Pi }{6}+2\Pi =\frac{13\Pi }{6}} as the limit on the right. I.e.:
\latex{\frac{5\Pi }{6}+2k\Pi \leq 2x\leq \frac{13\Pi }{6}+2k\Pi (k\in \Z).}
From here the solution of the inequality:
\latex{\frac{5\Pi }{12}+k\Pi \leq x\leq \frac{13\Pi }{12}+k\Pi .}
Note: It can be seen that we considered the period too when formulating the condition for \latex{2x} , and that we calculated with it in the same way as with the given end-points of the interval. The final result obtained shows that \latex{\sin 2x} will be periodic not with the period \latex{2\Pi} but with the period \latex{\Pi}.
\latex{ y }
\latex{ x }
Figure 47
Example 6
For which real numbers is it true that \latex{\sin 2x\leq \cos x}?
Solution
By applying the relation relating to the double-angles the inequality canbe rewritten in the following form:
\latex{2\times \sin x\times \cos x\leq \cos x,}
\latex{0\leq \cos x\times (1-2\times \sin x).}
\latex{0\leq \cos x\times (1-2\times \sin x).}
By taking into account that a product is non-negative if and only if its factors have the same sign, the following two cases may occur:
- If \latex{\cos x\geq 0} and \latex{1-2\times \sin x\geq 0,} i.e. \latex{\cos x\geq 0} and \latex{\sin x\leq \frac{1}{2}.}
By representing the conditions in the unit circle (Figure 48):
\latex{-\frac{\Pi }{2}+k\times 2\Pi \leq x\leq \frac{\Pi }{2}+k\times 2\Pi}
\latex{-\frac{7\Pi }{6}+I\times 2\Pi \leq x\leq \frac{\Pi }{6}+I\times 2\Pi}
\latex{\frac{\Pi }{2}}
\latex{-\frac{\Pi }{2}}
\latex{-\frac{7\Pi }{6}}
\latex{\frac{1}{2}}
\latex{\frac{\Pi }{6}}
\latex{ y }
\latex{ x }
\latex{ y }
\latex{ x }
Figure 48
\latex{-\frac{\Pi }{2}+n\times 2\Pi \leq x\leq \frac{\Pi }{6}+n\times 2\Pi}
\latex{\frac{1}{2}}
\latex{\frac{\Pi }{6}}
\latex{-\frac{\Pi }{2}}
\latex{ x }
\latex{ y }
Figure 49
According to these:
\latex{-\frac{\Pi }{2}+k\times 2\Pi \leq x\leq \frac{\Pi }{2}} and \latex{-\frac{7\Pi }{6}+l\times 2\Pi \leq x\leq \frac{\Pi }{6}+l\times 2\Pi (k\in \Z).}
The intersection of the two intervals (Figure 49):
\latex{-\frac{\Pi }{2}+n\times 2\Pi \leq x\leq \frac{\Pi }{6}+n\times 2\Pi (n\in \Z).}
- If \latex{\cos x\leq 0} and \latex{1-2\times \sin x\leq 0}, i.e. \latex{\cos x\leq 0} and \latex{\sin x\geq \frac{1}{2}.}
By representing the conditions in the unit circle (Figure 50):
\latex{\frac{\Pi }{2}+k\times 2\Pi \leq x\leq \frac{3\Pi }{2}+k\times 2\Pi}
\latex{\frac{\Pi }{6}+I\times 2\Pi \leq x\leq \frac{5\Pi }{6}+I\times 2\Pi}
\latex{\frac{\Pi }{2}}
\latex{\frac{3\Pi }{2}}
\latex{\frac{5\Pi }{6}}
\latex{\frac{1}{2}}
\latex{\frac{\Pi }{6}}
\latex{ y }
\latex{ x }
\latex{ y }
\latex{ x }
Figure 50
\latex{\frac{5\Pi }{6}}
\latex{\frac{1}{2}}
\latex{\frac{\Pi }{2}}
\latex{\frac{\Pi }{2}+n\times 2\Pi \leq x\leq \frac{5\Pi }{6}+n\times 2\Pi}
\latex{ x }
\latex{ y }
Figure 51
According to these:
\latex{\frac{\Pi }{2}+k\times 2\Pi \leq x\leq \frac{3\Pi }{2}+k\times 2\Pi} and \latex{\frac{\Pi }{6}+l\times 2\Pi \leq x\leq \frac{5\Pi }{6}+l\times 2\Pi (k,l\in \Z).}
The intersection of the two intervals (Figure 51):
\latex{\frac{\Pi }{2}+n\times 2\Pi \leq x\leq \frac{5\Pi }{6}+n\times 2\Pi (n\in \Z).}
To summarise: the solution of the inequality (Figure 52):
\latex{\frac{\Pi }{2}+n\times 2\Pi \leq x\leq \frac{5\Pi }{6}+n\times 2\Pi} or \latex{-\frac{\Pi }{2}+n\times 2\Pi \leq x\leq \frac{\Pi }{6}+n\times 2\Pi (n\in \Z).}
\latex{\frac{5\Pi }{6}}
\latex{\frac{1}{2}}
\latex{\frac{\Pi }{2}}
\latex{\frac{\Pi }{6}}
\latex{-\frac{\Pi }{2}}
\latex{ x }
\latex{ y }
Figure 52
Exercises
{{exercise_number}}. Solve the following equations on the set of real numbers:
- \latex{|\sin x|=\frac{\sqrt{3} }{2};}
- \latex{\cos ^{2}x=\frac{1}{4};}
- \latex{\tan ^{2}x=1;}
- \latex{|\cot x|=\sqrt{3}}
{{exercise_number}}. Solve the following equations:
- \latex{\cos x=\sin (x-\frac{\Pi }{4} );}
- \latex{\sin x=\cos (x+\frac{\Pi }{3} );}
- \latex{\sin (x-\frac{\Pi }{3} )=\cos (\frac{\Pi }{6}-x ).}
{{exercise_number}}. Find the solutions of the following equations:
- \latex{\sin x-\sqrt{3}\times \cos x=1}
- \latex{\sin x+\cos x=\sqrt{2};}
- \latex{\sin x-\cos x=1}
{{exercise_number}}. For which real numbers is the following equality true?
\latex{\frac{2}{\sin 2x}=\tan x+\cot x.}
{{exercise_number}}. Solve the following equations on the set of real numbers:
- \latex{2\times \sin ^{2}x-3\times \sin x+1=0;}
- \latex{4\times \sin x\times \cos ^{2}x=3;}
- \latex{4\times \sin x\times \cos x\times \cos 2x=1.}
{{exercise_number}}. For which real numbers are the following inequalities true?
- \latex{\sin x\geq \frac{\sqrt{3} }{2};}
- \latex{\cos 2x\lt \frac{1}{2};}
- \latex{2\times \sin x\times \cos x\leq \frac{\sqrt{3} }{2}.}
{{exercise_number}}. Solve the following inequalities:
- \latex{\sin 2x\leq -\cos x;}
- \latex{\sin 2x\geq 2\times \cos x;}
- \latex{\cos 2x\leq \sin x.}
