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

1
by namku

2015-04-14T19:14:20+05:30

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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.

what is sin3pi/8
Sin 3 π/8 = Sin 3 * 180/8 = Sin 67.5 deg = 1.207 = (1+√2)/2
Sin -π/8 = - Sin 22.5 deg = - 0.2071 = (1 - √2)/2
if you have understood the three ways of solving the problem, i am happy. it took a lot of time and effort. I suppose u realize that. have u got benefited ?
yes ofc i said thanks right ?