Find the values of m for which given equation has real roots sin 2 ⁡ x − ( m − 3 ) sin ⁡ x + m = 0So I started by first satisfying D &gt;= 0 and found that m should range from ( − ∞ , 1 ] ∪ [ 9 , ∞ )However after this, I found roots by quadratic formula as sin ⁡ x = ( m − 3 ) / 2 ± ( m − 1 ) ( m − 9 ) / 4 However this value should be within range of sin function and I'm unable to solve the inequality so formed for values of m.How to solve the inequality( actually the ± term is confusing me)?

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Find the values of m for which given equation has real roots      sin    2    ⁡  x  −  (  m  −  3  )  sin  ⁡  x  +  m  =  0So I started by first satisfying   D  &amp;gt;=  0 and found that m should range from   (  −  ∞  ,  1  ]  ∪  [  9  ,  ∞  )However after this, I found roots by quadratic formula as  sin  ⁡  x  =  (  m  −  3  )      /    2  ±      (    m    −    1    )    (    m    −    9    )          /        4  However this value should be within range of sin function and I&#039;m unable to solve the inequality so formed for values of m.How to solve the inequality( actually the   ± term is confusing me)?

Alannah Yang · Accepted Answer

Step 1When   m  =  1, both roots of the quadratic are −1 so a solution for x exists. When   m  ∈  (  −  ∞  ,  1  ), the smaller root       r    1    =  (  m  −  3  )      /    2  −      1    2        (    m    −    1    )    (    m    −    9    )    &amp;lt;  −  1 as   (  m  −  3  )      /    2  &amp;lt;  −  1. So we should enforce that the larger root       r    2    =  (  m  −  3  )      /    2  +      1    2        (    m    −    1    )    (    m    −    9    )    ∈  [  −  1  ,  1  ], i.e.                                  −        1        ≤        (        m        −        3        )                  /                2        +                  1          2                          (          1          −          m          )          (          9          −          m          )                ≤        1                                                  ⟺                1        −        m        ≤                  (          1          −          m          )          (          9          −          m          )                ≤        5        −        m            Note that the first inequality is always true since   9  −  m  ≥  1  −  m  ≥  0 and hence   (  1  −  m  )  (  9  −  m  )  ≥  (  1  −  m      )    2  . The second inequality on squaring gives      m    2    −  10  m  +  9  ≤      m    2    +  25  −  10  mwhich is true. So all values of m≤1 will work.Step 2When   m  ≥  9, the larger root       r    2    ≥  3. So we must enforce   −  1  ≤      r    1    ≤  1, i.e.                                  −        1        ≤        (        m        −        3        )                  /                2        −                  1          2                          (          m          −          1          )          (          m          −          9          )                ≤        1                                                  ⟺                1        −        m        ≤        −                  (          m          −          1          )          (          m          −          9          )                ≤        5        −        m                                                  ⟺                m        −        5        ≤                  (          m          −          1          )          (          m          −          9          )                ≤        m        −        1            Correspondingly note that the second inequality is always true this time around. Squaring the first inequality would yield   25  ≤  9, which is never true. So no   m  ≥  9 works.Our final answer is   m  ≤  1.     ◼

Find the values of m for which given equation has real roots sin^2 x-(m-3)sin x+m=0

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