SOLUTION: Find x, 4^sin²x + 4^cos²x = 3√(2)

Question 1208089: Find x,
4^sin²x + 4^cos²x = 3√(2)
Answer by ikleyn(52832) (Show Source): You can put this solution on YOUR website!
.
Find x, + = .
~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The original equation is

     +  = .     (1)


Use  = .  


Then equation (1) becomes

     +  = .    (2)


Introduce new variable t = .  Then equation (2) takes the form

    t +  = .    (3)



Reduce it to the standard form quadratic equation

    t^2 + 4 = ,

    t^2 -  + 4 = 0.    (4)


Use the quadratic formula to find the roots

     =  =  = .



So, equation (4) has two roots.  

One root is       =  =  = .

Another root is   =  =  = .



So, further we consider two cases.



Case 1.   = .  


         Then    =  =  =  = .


         It implies  sin(x) = +/-  = +/- .


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




Case 2.   = .  


         Then    =  =  =  = .


         It implies  sin(x) = +/-  = +/- .


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



ANSWER.  The solutions to given equation are  these four infinite sets of real numbers

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

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

Solved.

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