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

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Question 1208089: Find x,
4^sin²x + 4^cos²x = 3√(2)
Answer by ikleyn(52818) About Me  (Show Source):
You can put this solution on YOUR website!
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Find x,   4%5Esin%5E2%28x%29 + 4%5Ecos%5E2%28x%29 = 3%2Asqrt%282%29.
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The original equation is

    4%5Esin%5E2%28x%29 + 4%5Ecos%5E2%28x%29 = 3%2Asqrt%282%29.     (1)


Use cos%5E2%28x%29 = 1-sin%5E2%28x%29.  


Then equation (1) becomes

    4%5Esin%5E2%28x%29 + 4%2F4%5Esin%5E2%28x%29 = 3%2Asqrt%282%29.    (2)


Introduce new variable t = 4%5Esin%5E2%28x%29.  Then equation (2) takes the form

    t + 4%2Ft = 3%2Asqrt%282%29.    (3)



Reduce it to the standard form quadratic equation

    t^2 + 4 = 3t%2Asqrt%282%29,

    t^2 - 3t%2Asqrt%282%29 + 4 = 0.    (4)


Use the quadratic formula to find the roots

    t%5B1%2C2%5D = %283%2Asqrt%282%29+%2B-+sqrt%28%283%2Asqrt%282%29%29%5E2+-+4%2A4%29%29%2F2 = %283%2Asqrt%282%29+%2B-+sqrt%2818-16%29%29%2F2 = %283%2Asqrt%282%29+%2B-+sqrt%282%29%29%2F2.



So, equation (4) has two roots.  

One root is      t%5B1%5D = %283%2Asqrt%282%29%2Bsqrt%282%29%29%2F2 = %284%2Asqrt%282%29%29%2F2 = 2%2Asqrt%282%29.

Another root is  t%5B2%5D = %283%2Asqrt%282%29-sqrt%282%29%29%2F2 = %282%2Asqrt%282%29%29%2F2 = sqrt%282%29.



So, further we consider two cases.



Case 1.  4%5Esin%5E2%28x%29 = 2%2Asqrt%282%29.  


         Then   sin%5E2%28x%29 = log%284%2C2%2Asqrt%282%29%29 = %281%2F2%29%2Alog%282%2C2%2Asqrt%282%29%29 = %281%2F2%29%2A%283%2F2%29 = 3%2F4.


         It implies  sin(x) = +/- sqrt%283%2F4%29 = +/- sqrt%283%29%2F2.


         Hence,  x = pi%2F3+%2B+k%2Api,  k = 0, +/-1, +/-2, . . .  or  x = -pi%2F3+%2B+k%2Api,  k = 0, +/-1, +/-2, . . .




Case 2.  4%5Esin%5E2%28x%29 = sqrt%282%29.  


         Then   sin%5E2%28x%29 = log%284%2Csqrt%282%29%29 = %281%2F2%29%2Alog%282%2Csqrt%282%29%29 = %281%2F2%29%2A%281%2F2%29 = 1%2F4.


         It implies  sin(x) = +/- sqrt%281%2F4%29 = +/- 1%2F2.


         Hence,  x = pi%2F6+%2B+k%2Api,  k = 0, +/-1, +/-2, . . .  or  x = -pi%2F6+%2B+k%2Api,  k = 0, +/-1, +/-2, . . .



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

         x = pi%2F6+%2B+k%2Api,  k = 0, +/-1, +/-2, . . .  or  x = -pi%2F6+%2B+k%2Api,  k = 0, +/-1, +/-2, . . .

     or  x = pi%2F3+%2B+k%2Api,  k = 0, +/-1, +/-2, . . .  or  x = -pi%2F3+%2B+k%2Api,  k = 0, +/-1, +/-2, . . .

Solved.