Find parameter so trigonometric equation has real solutionFind the parameter m so the equation has real solution: ( sin ⁡ x − m ) 2 + ( 2 m sin ⁡ x − 1 ) 2 = 0I tried rewriting it like this: sin 2 ⁡ x ( 1 + 4 m 2 ) − 6 m ⋅ sin ⁡ x + 1 + m 2 = 0And imposing that the discriminant is ≥ 0, the sum of the roots ∈ [ − 2 , 2 ] and the product of the roots ∈ [ − 1 , 1 ] (all that because the roots should be in [ − 1 , 1 ] ). Still, the answer is { − 2 2 , 2 2 } .

Question

Find parameter so trigonometric equation has real solutionFind the parameter   m so the equation has real solution:  (  sin  ⁡  x  −  m      )    2    +  (  2  m  sin  ⁡  x  −  1      )    2    =  0I tried rewriting it like this:      sin    2    ⁡  x  (  1  +  4      m    2    )  −  6  m  ⋅  sin  ⁡  x  +  1  +      m    2    =  0And imposing that the discriminant is   ≥  0, the sum of the roots   ∈  [  −  2  ,  2  ] and the product of the roots   ∈  [  −  1  ,  1  ] (all that because the roots should be in   [  −  1  ,  1  ]  ). Still, the answer is       {    −                  2            2        ,                  2            2        }  .

Kristen Garza · Accepted Answer

For any real number   α, we have       α    2    ≥  0 thus we must have both   (  sin  ⁡  x  −  m      )    2    =  0 and   (  2  m  sin  ⁡  x  −  1      )    2    =  0So we have   sin  ⁡  x  =  m and   sin  ⁡  x  =      1          2      m      This gives   m  =      1          2      m      And   m  =      1          2       or   m  =            −      1              2

Find parameter so trigonometric equation has real solution Find the parameter m so the equation has real solution: (sinx−m)^2+(2 m sin x−1)^2=0 I tried rewriting it like this: sin^2 x(1+4m^2)−6m*sinx+1+m^2=0

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