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Trigonometric Equations

  1. Sep 2, 2010 #1
    1. The problem statement, all variables and given/known data

    Solve (1+cosx)sin^2x = x^2 + 1/(x^2)

    2. Relevant equations



    3. The attempt at a solution

    RHS is always greater than or equal to 2
    LHS is always less than or equal to 2 as
    1+cosx is always less than or equal to 2 and sin^2x is always less than or equal to 1

    so there is a solution when LHS=RHS=2
    but my book says there is no solution as LHS is always less than 2 (not equal to 2)

    How?
     
  2. jcsd
  3. Sep 2, 2010 #2
    You’re off to a good start.

    Let y = cosx and z = sin^2(x), both of these are inclusively bound between 0 and 1, Your LFH then becomes (1+y)z, where 0 <= y <= 1; 0 <= z <= 1. So what value does y and z need to be for the LHS to equal 2?

    So this puts your equation under the constraints of cosx = 1; sin^2(x) = 1; x^2 + x^-2 = 2

    Note sin^2(x) = 1 => sin(x) = +/-1

    When does cos(x) = 1? When does sin(x) = =/-1

    Can you ever meet both of these constraints with the same value of x?
     
  4. Sep 2, 2010 #3
    Thanks....I got it :)
     
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