Lesson Solving trigonometric equations that include a multiple of an unknown as an argument

Solving trigonometric equations that include a multiple of an unknown as an argument

Problem 1

Your starting equation is

     = 1


Transform it step by step

     = 1,

    1 + 2*sin(x)*cos(x) = 1,

    sin(2x) = 0.


Hence, 2x should be a multiple of .


Since x should be in interval [,),  2x should be in interval [,)

    0 <= 2x < .


There are 4 possible values for 2x: 0,  ,    and  .


It gives 4 possible values for x:  0, ,    and  .


It is easy to check that all these 4 values satisfy given equation.


ANSWER.  Given equation has 4 solutions in the given interval:  0,  ,  and .

Problem 2

Your starting equation is

    sec^2 x + 2tan x = 0.


Transform it equivalently this way

     = 0.


Now multiply both sides by cos(x).  Since cos(x) =/=0 by the meaning of the original equation,
we can do it secure in the equation's domain. We get then  

    1 + 2sin(x)*cos(x) = 0.

    1 + sin(2x) = 0,

    sin(2x) = -1


Hence, 

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


We take only two values  2x =   (k=0)  and  2x =   (k=1),
since other values produce values of x out of the given interval.


ANSWER.  The given equation has two roots in the interval  [,).

         They are , or 135°,  and  , or 315°.

Problem 3

Our starting equation is

    sec(4t) = .


Transform it equivalently this way

    cos(4t) = 

    4t =   OR  ,  'k'  is any integer.

Hence,

    t =   OR  t = ,  'k'  is any integer.    ANSWER

Problem 4

The starting equation is

    cos(5x) = -1.


The table solution for cosine is  

    5x = π, 3π, 5π, 7π, 9π, . . . (all odd integer multiples of π).


It implies

    x = , ,   = π,   ,  , . . . 


For greater or smaller values of 5x,  values of x will be out of the given interval - so, we do not consider them.


Thus, there are 5 solutions to the given equation in interval 0 <= x < 2π.


The smallest solution is  .  The greatest solution is  .  The number of solutions is 5.  ANSWER

Problem 5

tan(x/8) + √3 = 0,    (1)


tan(x/8) = -√3.       (2)


We are looking for possible solutions 'x' in interval [,). 


So, x/8 should be in interval [0,), which is the same as interval [0,). 


But in this interval from 0 to , function tan is always non-negative.


Hence, equation (2) with the negative right side HAS NO solutions in this interval.


It implies that the original equation has no solution/solutions in the given interval [0,2pi).     ANSWER

Problem 6

Use identity:  cos(x+y) = cos(x)*cos(y) - sin(x)*sin(y)

cos(2x)* cos(x) − sin(2x)* sin( x) = square root of 2 over 2

cos(2x+x) = √2/2

2x+x = π/4 + 2k*π;     7π/4 + 2k*π.

  3x = π/4 + 2k*π;     7π/4 + 2k*π.


    For 3x, you must add the periods 2π and 4π for cosine.
    For 3x, it will produce geometrically the same angle; 
    but for 'x' it will produce new angles that you will miss otherwise.

    You should no add more hire periods for 3x, since it will not produce 
    new angles for x and will lead you out of the given interval.


Now consider two cases.


Case (a).  3x = ,  ,  .

            then  x = ,   =  = ,   = .



Case (b).  3x = ,  ,  .

            then  x = ,   =  = ,   = .



ANSWER.  The given equation has 6 solutions in the given interval

         ,  ,  ,  ,  ,  .

Problem 7

Equation

    cos(2x+20°) = sin(3x−10°)    (1)

implies

    (2x+20°) + (3x-10°) = = 90° + k*360°, k = 0, +/-1, +/-2, . . . (2)


because both functions tan and cot are periodical with period  .


So, we can rewrite equation (2) in the form

    5x = 80° + k*360°, k = 0, 1, 2, 3, 4.


We can limit ourselves with 5 values of k, because for other values of 'k', the angles 'x' will be geometrically the same,
but formally will be outside of the interval [0°,360°).


For these 5 values of 'k', we will have 5 different values of x in the interval [0°,360°)

    x = 16°,  16°+72° = 88°,  16°+144° = 160°,  16°+216° = 232°,  16°+288° = 304°.


ANSWER. Taking into account the periodicity of the left side and the right side of the given equation, 

it is natural to look for solutions 'x' in the interval [0°,360°).

In this interval, the given equation has 5 (five) different solutions 16°, 88°, 160°, 232° and 302°.

Problem 8

Equation

    tan(x) = cot(4x+20°)    (1)

implies

    x + (4x+20°) = 90° + k*180°, k = 0, +/-1, +/-2, . . . (2)


because both functions tan and cot are periodical with period .


So, we can rewrite equation (2) in the form

    5x = 70° + k*180°, k = 0, 1, 2, 3, 4.


We can limit ourselves with 5 values of k, because for other values of 'k', the angles 'x' will be geometrically the same,
but formally will be outside of the interval [0°,180°).


For these 5 values of 'k', we will have 5 different values of x in the interval [0°,180°)

    x = 14°,  14°+36° = 50°,  14°+72° = 86°,  14°+108° = 122°,  14°+144° = 158°.


ANSWER. Taking into account the periodicity of the left side and the right side of the given equation, 

it is natural to look for solutions 'x' in the interval [0°,180°).

In this interval, the given equation has 5 (five) different solutions 14°, 50°, 86°, 122° and 158°.