SOLUTION: solve for x: {{{(cos(x)+sin(x))^(sin(2x)+1)=2}}}

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Question 1151776: solve for x:
%28cos%28x%29%2Bsin%28x%29%29%5E%28sin%282x%29%2B1%29=2

Answer by ikleyn(52756) About Me  (Show Source):
You can put this solution on YOUR website!
.

            The point is that  2  is the maximal value of the given function.

            It can be proved by standard method of  Calculus:

                    by taking the derivative,  equating it to zero,  and then solving the associated equation.

            In this post I will show you another way,  which quickly leads to the answer.


(1)  First notice that  cos(x)+sin(x)  has the maximum value of  sqrt%282%29  at x = pi%2F4.      // Every Calculus student must know it (!)


(2)  At this value of x,  sin(2x) = 1  and,  therefore,  sin(2x)+1 = 2.  It is maximum value of  sin(2x)+1.


(3)  Therefore,  at  x= pi%2F4  %28cos%28x%29%2Bsin%28x%29%29%5E%28sin%282x%29%2B1%29 = %28sqrt%282%29%29%5E2 = 2.


(4)  At all other values of x  in the interval  [0,2pi),  cos(x) + sin(x)  is less than  sqrt%282%29

     and  sin(2x)+1  is less than  2;  therefore, the entire function  %28cos%28x%29%2Bsin%28x%29%29%5E%28sin%282x%29%2B1%29  is strictly less than 2.


(5)  Thus, the only solution to the original equation in the interval  [0,2pi)  is  x= pi%2F4.

Solved.