Question

the equation sinx(sinx+cosx)=p has real roots . find the interval of p

Answer 1

Sin x ( sin x + cos x ) = p
 Sin² x + Sin x cos x = p
(1 - Cos 2 x ) / 2 +  1/2 Sin 2 x  = p
 1 - Cos 2 x + Sin 2 x  = 2 p
Sin 2x - Cos 2 x = 2p - 1
Sin 2x - SIn (π/2 - 2x) = 2p - 1

the  rule  Sin A - Sin B =  2 Sin (A - B)/2 Cos (A+B)/2 

 So    2 Sin ( 2x - π/4)  Cos π/4 =  2p -1
           Sin (2 x - π/4)  =  (2p -1) / √2
       Sin of an angle is always between  + 1 and -1.

   So    -1 ≤  (2p -1)/√2  ≤  1
                   2p -1  >= - √2      =>    p >=  (1-√2)/2
          and  2 p - 1 <=  √2      =>     p <=  (1+√2)/2


The limits for p , or the interval for p  is  :    0  ≤  p  ≤ 1

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Sin x ( sin x + cos x ) = p
Sin x (sin x + Sin (90 -x)  )  = p
           we use the  rule  Sin A + Sin B =  2 Sin (A + B)/2 Cos (A-B)/2 

  Sin x [ 2 Sin 45⁰ Cos (x - 45⁰) ]= p

     2  Sin x  Cos (x - 45°) = √2 p

      Sin (2 x - 45°) + Sin  45° = √2 p
       Sin (2x - 45°)  =  √2 p - 1/√2 = (2p -1)/√2
   As Sine of an angle is always between -1 and 1,  we get that,
                   2p -1  >= - √2      =>    p >=  (1-√2)/2
          and  2 p - 1 <=  √2      =>     p <=  (1+√2)/2
 
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if you know differentiation then:

Sin x ( sin x + cos x ) = p
  Sin² x  + sin x  Cos x = p
  Sin² x  + 1/2  Sin 2x  = p

dp / dx = deriviate wrt x =  2 Sin x + Cos 2 x 
  make the derivative = 0  to find maximum or minimum.
 
   Hence,  2 Sin x + 1 - 2 Sin² x = 0
                 Sin² x - Sin x - 1/2 = 0

  this is a quadratic equation :    Sin x = [ 1 + -  √(1+2) ] / 2
       Sin x =  (1 + √3 )/2  or  (1 - √3)/2
                = only  (1 - √3)/2  as the other one is more than 1. so x is not real.
                   = -0.366

   Then  cos² x  = 1 - sin² x = 1 -  1/4 [ 1 + 3 - 2√3 ]
                         = √3/2
             Cos x = √(√3/2) = 0.9306
          
substitute the values in the expression to get the max  or min value.