Lesson Miscellaneous problems on solving trigonometric equations

Miscellaneous problems on solving trigonometric equations

Problem 1

The original equation is 

    sin(3x) + cos(3x) = -1.    (1)   


Square its both sides. You will get

     = .   (2)


From the other side, there is an identity

     == .   (3)


Comparing (2) and (3), you get

    2*sin(3x)*cos(3x) = 0,   or  sin(3x)*cos(3x) = 0.    (4)


Equation (4) splits in two independent equations


    1)  sin(3x) = 0  --->  3x = ,   k = 0. +/-1. +/-2, . . . 
                     --->   x = ,  k = 0, +/-1. +/-2, . . .       (5)


    2)  cos(3x) = 0  --->  3x = ,   k = 0. +/-1. +/-2, . . . 
                     --->   x = ,  k = 0. +/-1. +/-2, . . .   (6)


Now we should check which of the found values (5), (6) satisfy the original equation.


Of the set (5), all x satisfy sin(3x) = 0. Hence, only those of (5) satisfy the original equation where cos(3x) = -1.
They are  3x = ,    n = 0, +/-1. +/-2, . . . ,  or 

           x = ,  n = 0, +/-1. +/-2, . . . ,           (5').


Of the set (6), all x satisfy cos(3x) = 0. Hence, only those of (6) satisfy the original equation where sin(3x) = -1.
They are  3x = ,   n = 0, +/-1. +/-2, . . . ,  or 

           x = ,  n = 0, +/-1. +/-2, . . . ,           (6'). 


Answer.  The union of the sets (5') and (6') is the solution of the original equation. 

Problem 2

Since  cos(6x) = cos(x), it implies one of two possibilities:


    (1)  EITHER  6x =  x + 360n  degrees

    (2)   OR     6x = -x + 360n  degrees.



From (1), we have  6x-x = 360n;  5x = 360n;   y = 0,  360/5 = 72,  (360/5)*2, . . . , (360/5)*5.

          Taking into account the restriction on the range, the only possible solutions are 0°, 72°,  144°, 216°, 288°.



From (2), we have  6x+x = 360n;  7x = 360n;   x = 0,  360/7, (360/7)*2,  (360/7)*3,  (360/7)*4,  (360/7)*5,  (360/7)*6.  

          Taking into account the restriction on the range, the only possible solutions are 0°  and  360/7 degrees.


It gives the ANSWER : the solutions for x  are  0°, 72°,  144°, 216°, 288°,

                                                      360/7, (360/7)*2,  (360/7)*3,  (360/7)*4,  (360/7)*5,  (360/7)*6  degrees.

Problem 3

Your starting equation is 

     −  = 0    for 0 <= x < .


It is the same as  (is equivalent to)

     =    for 0 <= x < .


Since tan is a monotonic periodic function with the period of  π, it means that

     -  = ,   where  k = 0, +/-1, +/-2, . . . 

or

    2x = ,    where  k = 0, +/-1, +/-2, . . . 



Having x in the interval  0 <= x < ,  it means that  

    x = 0  or  x = .        ANSWER

Problem 4

Your starting equation is

    2cos^2(x) + sin(20°) = 1.      (1)


Notice that sin(20°) = cos(90°-20°) = cos(70°).


Therefore, you can re-write the original equation in the form

    2cos^2(x) = 1 - cos(70°),  and further as  cos^2(x) = .      (2)


Next, compare it with the formula for sine of the half angle

     = .


Taking a = 70°, you see from (2) that

     = .       (3)


From (2) and (3), we have

     = .      (4)


Now use  sin(35°) = cos(90°-35°) = cos(55°).  Then equation (4) takes the form

     = .      (5)


It gives for x the following possibilities  

    x = 55°,  180°-55° = 125°,  55°+180° = 235°  and  360°-55° = 305°.


Since we are looking for the solutions in the interval 0° < x < 180° only, it leaves only two possible values.


ANSWER.  In the given interval, the solutions are  x= 55°  and  x= 125°.


CHECK.   cos(55°) = 0.573576;  sin(20°) = 0.342020;  2*0.573576^2 + 0.342020 = 0.999999.   ! Good precision, correct !