Trigonometric equation with two functions and a parameterThe problem: For which values of a the equation a . s i n x . c o s x = s i n x − c o s x has exactly 2 different roots in the interval [ 0 ; π ].So clearly 0 , π 2 , π are not an answer. I divide by sinx.cosx and get f ( x ) = 1 c o s x − 1 s i n x = a. I look at the interval ( 0 ; π 2 ) cos x is decreasing so 1 c o s x is increasing same for sin x is increasing and − 1 s i n x is also increasing. lim x → 0 f ( x ) = + ∞ and lim x → π 2 f ( x ) = − ∞ and f(x) is continuous so we have exactly 1 root in the interval ( 0 ; π 2 ) for every a.My question: I guess I have to find for what value of a the function has 1 root in the interval ( π 2 ; π ) how do I do that and is there an easier way to solve the whole problem?

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Trigonometric equation with two functions and a parameterThe problem: For which values of a the equation   a  .  s  i  n  x  .  c  o  s  x  =  s  i  n  x  −  c  o  s  x has exactly 2 different roots in the interval   [  0  ;  π  ].So clearly   0  ,      π    2    ,  π are not an answer. I divide by sinx.cosx and get   f  (  x  )  =      1          c      o      s      x        −      1          s      i      n      x        =  a. I look at the interval   (  0  ;      π    2    ) cos x is decreasing so       1          c      o      s      x       is increasing same for sin x is increasing and             −      1              s      i      n      x       is also increasing.            lim              x        →        0              f    (    x    )    =    +    ∞   and             lim              x        →                  π          2                      f    (    x    )    =    −    ∞   and f(x) is continuous so we have exactly 1 root in the interval   (  0  ;      π    2    ) for every a.My question: I guess I have to find for what value of a the function has 1 root in the interval   (      π    2    ;  π  ) how do I do that and is there an easier way to solve the whole problem?

sekanta2b · Accepted Answer

Step 1It may be useful to look at the graphs of the given functions. The given equation can be converted to the following,   a  sin  ⁡      2    x    =  2      2    sin  ⁡      (    x    −          π      4        )  Step 2Now sketch the graphs and observe that you will have two solutions iff   a  =  −  2      2  . Here one solution will lie in   (  0  ,  π      /    2  ) where the graphs will intersect transversely, whereas one more solution in   (  π      /    2  ,  π  ) where the graphs will &#039;touch&#039; each other tangentially. Now you can play with the values of a and see that the number of solutions will increase and decrease if you change a slightly.

Taliyah Reyes · Accepted Answer

Step 1Let   f  (  x  )  =      1          c      o      s      x        −      1          s      i      n      x        ,  x  ∈  (  0  ,      π    2    ). Then f is strictly increasing and             lim              x        →        0              f    (    x    )    =    −    ∞    ,                  lim                  x          →                      π            2                              f      (      x      )      =      +      ∞      therefore the equation   f  (  x  )  =  a has a unique solution on   (  0  ,      π    2    )Step 2For   x  ∈  (      π    2    ,  π  ), then             lim              x        →                              π            2                                f    (    x    )    =    −    ∞   and             lim              x        →                  π                      f    (    x    )    =    −    ∞  . To have a unique solution in   (      π    2    ,  π  ), we must take a the maximum value of f. Can you take it from here?

For which values of a the equation a.sin x. cos x=sin x-cos x has exactly 2 different roots in the interval [0 ;pi].

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