Lesson Solving equation by analyzing function

Solving equation by analyzing function

Problem 1

Our starting equation is 

     = cos(x).    (1)


Let's consider function    first, which is under the outer square root of the left side.


Inside the first quadrant, QI, when  0 < x < ,  both sin(x) and cos(x) are positive.

Therefore, the sum sin(x) + cos(x) is greater than sin(x): 

    sin(x) + cos(x) > sin(x).         (2)


Square root is a monotonic function of its argument; therefore, from inequality (1) we have

     > .    (3)


Since inside the first quadrant, QI, sin(x) is less than 1, we have

     > sin(x).                (4)


So, combining (3) and (4), we have

    sin(x) < .


It means that inside QI the expression under the outer square root is negative;
so, left side of the given equation is not defined.


Let's inspect endpoints of QI, angles x = 0 and x = .


At x = 0, left side of the given equation is not defined.

At x = , the given equation is valid, so x =  is a solution.


Thus we found that in the first quadrant, 0 <= x <= , the only solution is  x = ,   
where both sides of the given equation are defined and the equation is valid.



In the second quadrant,   < x <= ,  the left side of the given equation 

is EITHER positive OR not defined, while the right side is always negative, so in the domain

 < x <=  our given equation has no solutions.



In QIII and QIV, where   < x < ,  

    (a) sin(x) is negative 

and the term

    (b)   is either negative or not defined.


Therefore, the left side of the given equation is not defined at   < x  < .


So, in the union  { QIII U QIV ) there is no solution to equation (1), at all.


Thus we found that the only solution to equation (1) in the interval  [,)  is  x = .


If you want to get the GENERAL solution for any real x, then use the fact that both
left side and right side of equation (1) are periodic functions of x with the period .


Hence, the general solution to equation (1) is the set of values  ,  
where  'k'  is any integer  k = 0, +/-1, +/-2, . . .  and so on.


At this point, the problem is solved completely.