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B Trigonometric equation -- real roots

  1. Mar 3, 2017 #1
    The number of real roots of the equation

    $$2cos \left( \frac {x^2 + x} {6} \right)=2^x + 2^{-x}$$

    Answer options are : 0,1,2,∞

    My approach :

    range of cos function is [-1,1]
    thus the RHS of the equation belongs to [-2,2]
    So, we have
    -2 ≤ 2x + 2-x ≤ 2
    solving the right inequality, i got 2x = 1 and that satisfies the left part too
    therefore , x can take only one value = 0
    Now since this values agrees with the orginal trigonometric equation, we have 1 real root for this equation

    So answer is 1

    Is my approach/answer correct ?
     
  2. jcsd
  3. Mar 3, 2017 #2

    QuantumQuest

    User Avatar
    Gold Member

    It is correct. You can also start from the RHS and try to see if ##2^{x} + 2^{-x}## bounds some expression with known value.
     
  4. Mar 3, 2017 #3

    fresh_42

    Staff: Mentor

    Looks good so far, but how do you get from ##2^x +2^{-x}\leq 2## to ##2^x=1##?
    I only see ##-1 < x < 1## without pen and paper.
     
  5. Mar 3, 2017 #4
    @fresh_42
    ##2^x +\frac{1}{2^x}\leq 2##
    ##(2^x)^2 - 2*2^x + 1 \leq 0##
    ##(2^x - 1)^2 \leq 0##

    only the equality condition is feasible for real x
     
  6. Mar 3, 2017 #5

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Alternatively, you can see that x=0 is a solution to ##2^x + 2^{-x} = 2## and consider the derivatives to show that there are no further solutions. The quadratic equation is more elegant, however.
     
  7. Mar 3, 2017 #6

    pasmith

    User Avatar
    Homework Helper

    [itex]2^x + 2^{-x} = 2\cosh(x \ln 2) \geq 2[/itex].
     
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