Lesson Challenging problems on trigonometric equations

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Challenging problems on trigonometric equations


In this lesson you will find the solutions of these trigonometric equations:

    1.  sin(x) + sin(2x) + sin(3x) = 0.     2.  sin(x) + sin(3x) + sin(5x) = 0.     3.  cos(x) + cos(2x) + cos(3x) + cos(4x) = 0.     4.  sin(x) + sin(2x) + sin(3x) + sin(4x) = 0.

Problem 1

Solve an equation   sin(x) + sin(2x) + sin(3x) = 0   over the interval  [0,2pi).

Solution 

sin(x) + sin(2x) + sin(3x) = 0.     (1)


Apply the trigonometry formula  sin%28a%29+%2B+sin%28b%29 = 2%2Asin%28%28a%2Bb%29%2F2%29%2Acos%28%28a-b%29%2F2%29

     (see any serious textbook in trigonometry or the lessons 

         - FORMULAS FOR TRIGONOMETRIC FUNCTIONS
         - Addition and subtraction of trigonometric functions

      in this site) to the first and third addend in the left side of the original equation (1).   You will get

sin%28x%29%2Bsin%283x%29 = 2%2Asin%28%28x%2B3x%29%2F2%29%2Acos%28%28x-3x%29%2F2%29 = 2%2Asin%282x%29%2Acos%28-x%29 = 2%2Asin%282x%29%2Acos%28x%29.


Now, the equation (1) takes the form

2%2Asin%282x%29%2Acos%28x%29+%2B+sin%282x%29 = 0,   or

sin%282x%29%2A%282%2Acos%28x%29+%2B1%29 = 0.


This equation deploys in two independent equations


1.  sin(2x) = 0  --->  x = k%2Api, x = pi%2F2+%2B+k%2Api,  k = 0, =/-1, +/-2. . . . 


2.  2cos(x) + 1 = 0  --->  cos(x) = -1%2F2  --->  x = 2pi%2F3+%2B+2k%2Api,  x = 4pi%2F3+%2B+2k%2Api,  k = 0, =/-1, +/-2. . . . 


Answer.  The solutions are  a)  x = k%2Api, x = pi%2F2+%2B+k%2Api,  k = 0, =/-1, +/-2. . . .    and 

                            b)  x = 2pi%2F3+%2B+2k%2Api,  x = 4pi%2F3+%2B+2k%2Api,  k = 0, =/-1, +/-2. . . . 


    


    Plot y = sin(x) + sin(2x) + sin(3x)

Problem 2

Solve an equation   sin(x) + sin(3x) + sin(5x) = 0   over the interval  [0,2pi).

Solution 

sin(x) + sin(3x) + sin(5x) = 0.              (1)


Using the Trigonometry formula   ,                    (*)

you can transform    sin%28x%29+%2B+sin%285x%29 = 2%2Asin%283x%29%2Acos%28-2x%29 = 2%2Asin%283x%29%2Acos%282x%29.


Then the left side of the given equation takes the form 

sin%28x%29+%2B+sin%283x%29+%2B+sin%285x%29 = 2%2Asin%283x%29%2Acos%282x%29 + sin%283x%29 = sin%283x%29%2A%282cos%282x%29+%2B+1%29,

and the equation (1) takes the form 
 
sin%283x%29%2A%282%2Acos%282x%29+%2B+1%29  =  0.                      (2)


Equation (2) deploys in two independent equations:


1)  sin(3x) = 0,  which in the given interval has the solutions  x = 0, pi%2F3, 2pi%2F3, pi, 4pi%2F3, and 5pi%2F3.


2)  2*cos(2x) + 1 = 0,  which is the same as  cos(2x) = -1%2F2.

    In the given interval the last equation has the solutions

    x = %281%2F2%29%2A%282pi%2F3%29,  %281%2F2%29%2A%284pi%2F3%29,  %281%2F2%29%2A%282pi%2F3+%2B2pi%29,  %281%2F2%29%2A%284pi%2F3+%2B2pi%29,    or, which is the same,

    x = pi%2F3,  2pi%2F3,  4pi%2F3  and  5pi%2F3.


Answer.  The solutions of the equation (1) in the interval [0,2pi)  are  x = 0, pi%2F3, 2pi%2F3, pi, 4pi%2F3, and 5pi%2F3.


    


    Plot y = Sin(x) + Sin(3x) + Sin(5x)

Problem 3

Solve an equation   cos(x) + cos(2x) + cos(3x) + cos(4x) = 0   over the interval  [0,2pi).

Solution 

cos(x) + cos(2x) + cos(3x) + cos(4x) = 0.        (1)


Use the general formula of Trigonometry

cos%28a%29+%2B+cos%28b%29 = 2cos%28%28a%2Bb%29%2F2%29%2Acos%28%28a-b%29%2F2%29.                (2)


You have 

cos(x) + cos(4x) = 2%2Acos%28%28x%2B4x%29%2F2%29%2Acos%28%284x-x%29%2F2%29 = 2%2Acos%282.5x%29%2Acos%281.5x%29,

cos(2x) + cos(3x) = 2%2Acos%28%282x%2B3x%29%2F2%29%2Acos%28%283x-2x%29%2F2%29 = 2%2Acos%282.5x%29%2Acos%280.5x%29.


Therefore, the left side of the original equation is

cos(x) + cos(2x) + cos(3x) + cos(4x) = 2*cos(2.5x)*cos(1.5x) + 2*cos(2.5x)*cos(0.5x) = 2*cos(2.5x)*(cos(1.5x) + cos(0.5x)).


Hence, the original equation is equivalent to

2*cos(2.5x)*(cos(1.5x) + cos(0.5x)) = 0,    or,  canceling  the factor  2*cos(2.5x),

cos(1.5x) + cos(0.5x) = 0.                       (3)


Again, apply the formula (2) to the left side of (3). You will get an equivalent equation

2%2Acos%28x%29%2Acos%28x%2F2%29 = 0.                                (4)


Equation (4) deploys in two independent separate equations:


1.  cos(x) = 0  --->  x = pi%2F2+%2B+k%2Api,  k = 0, +/-1, +/-2, . . . 


2.  cos(x/2) = 0  --->  x%2F2 = pi%2F2+%2B+k%2Api,  k = 0, +/-1, +/-2, . . . ,  or

                        x = pi+%2B+2k%2Api = %282k%2B1%29%2Api,  k = 0, +/-1, +/-2, . . . 


From (1) and (2), in the given interval the original equation has the roots pi%2F2, pi, 3pi%2F2,  or  90°,  180°,  270°.


But these are not ALL the roots.

There is one more family of roots.

Do you remember I canceled the factor 2*cos(2.5x) ?

Of course, I must consider (and add to the solution set !) all the solutions of the equation

cos(2.5x) = 0.

They are  2.5x = pi%2F2+%2B+k%2Api,  k = 0, +/-1, +/-2, . . . 

or, which is the same,

%285x%29%2F2 = pi%2F2+%2B+k%2Api,  k = 0, +/-1, +/-2, . . . 

So, these additional solutions are x = pi%2F5+%2B+2k%2A%28pi%2F5%29 = %282k%2B1%29%2A%28pi%2F5%29,  k = 0, +/-1, +/-2, . . . 

The final answer is:  There are two families of solutions. 

                      One family pi%2F2, pi, 3pi%2F2,  or  90°,  180°,  270°.

                      The other family is  pi%2F5, 3pi%2F5, 5pi%2F5 = pi, 7pi%2F5, 9pi%2F5,  or  36°, 108°, 180° (repeating root), 252°, 324°.

Solved.

CHECK

Look into the plot of the left side of the original equation 

    


    Plot y = cos(x) + cos(2x) + cos(3x) + cos(4x)



Do you see 7 roots in the interval [0,2pi) ?

Problem 4

Solve an equation   sin(x) + sin(2x) + sin(3x) + sin(4x) = 0   over the interval  [0,2pi).

Solution 

sin(x) + sin(2x) + sin(3x) + sin(4x) = 0.        (1)


Use the general formula of Trigonometry

sin%28a%29+%2B+sin%28b%29 = 2sin%28%28a%2Bb%29%2F2%29%2Acos%28%28a-b%29%2F2%29.                (2)


You have 

sin(x) + sin(4x) = 2%2Asin%28%28x%2B4x%29%2F2%29%2Acos%28%284x-x%29%2F2%29 = 2%2Asin%282.5x%29%2Acos%281.5x%29,

sin(2x) + sin(3x) = 2%2Asin%28%282x%2B3x%29%2F2%29%2Acos%28%283x-2x%29%2F2%29 = 2%2Asin%282.5x%29%2Acos%280.5x%29.


Therefore, the left side of the original equation is

sin(x) + sin(2x) + sin(3x) + sin(4x) = 2*sin(2.5x)*cos(1.5x) + 2*sin(2.5x)*cos(0.5x) = 2*sin(2.5x)*(cos(1.5x) + cos(0.5x)).


Hence, the original equation is equivalent to

2*sin(2.5x)*(cos(1.5x) + cos(0.5x)) = 0,    or,  canceling  the factor  2*sin(2.5x),

cos(1.5x) + cos(0.5x) = 0.                       (3)


Next, apply another general formula of Trigonometry

cos%28a%29+%2B+cos%28b%29 = 2cos%28%28a%2Bb%29%2F2%29%2Acos%28%28a-b%29%2F2%29.                (4)


Then the equation (3) becomes

2%2Acos%28x%29%2Acos%28x%2F2%29 = 0.                                (5)


Equation (5) deploys in two independent separate equations:


1.  cos(x) = 0  --->  x = pi%2F2+%2B+k%2Api,  k = 0, +/-1, +/-2, . . . 


2.  cos(x/2) = 0  --->  x%2F2 = pi%2F2+%2B+k%2Api,  k = 0, +/-1, +/-2, . . . ,  or

                        x = pi+%2B+2k%2Api = %282k%2B1%29%2Api,  k = 0, +/-1, +/-2, . . . 


From (1) and (2), in the given interval the original equation has the roots pi%2F2, pi, 3pi%2F2,  or  90°,  180°,  270°.


But these are not ALL the roots.

There is one more family of roots.

Do you remember I canceled the factor 2*sin(2.5x) ?

Of course, I must consider (and add to the solution set !) all the solutions of the equation

sin(2.5x) = 0.

They are  2.5x = k%2Api,  k = 0, +/-1, +/-2, . . . 

or, which is the same,

%285x%29%2F2 = k%2Api,  k = 0, +/-1, +/-2, . . . 

So, these additional solutions are x = 0, %282k%2Api%29%2F5, %284k%2Api%29%2F5, %286k%2Api%29%2F5, %288k%2Api%29%2F5,  k = 0, +/-1, +/-2, . . . 

The final answer is:  There are two families of solutions in the given interval. 

                      One family is  pi%2F2,  pi,  and  3pi%2F2,  or  90°, 180° and 270°.

                      The other family is  0,  %282pi%29%2F5,  %284pi%29%2F5,  %286pi%29%2F5,  %288pi%29%2F5,  or  0°, 72°, 144°, 216°, 288°.

Solved.

CHECK

Look into the plot of the left side of the original equation 

    


    Plot y = sin(x) + sin(2x) + sin(3x) + sin(4x)



Do you see 8 roots in the interval [0,2pi) ?


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Use this file/link  ALGEBRA-II - YOUR ONLINE TEXTBOOK  to navigate over all topics and lessons of the online textbook  ALGEBRA-II.

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