# Plz solve this in detail::If esinx - e-sinx - 4 = 0, then the no. of real solutions of x is.....??? plz explain!And also solve this:the greatest value of 1+ 12sinx - 9sin2x........????

Chetan Mandayam Nayakar
312 Points
12 years ago

let esinx=y, y-(1/y)-4=0, y^2 -4y-1=0, y=2±√8=esinx, -1≤sinx≤1, sinx=ln(y)=ln(2±√8), (2-√8)<0, so it is ruled out.

ln(2+√8)>1, so this is also ruled out.Thus, there are no real solutions.

1+12sinx-9sin2x=-(3sinx-2)2+5,it is greatest when sinx=(2/3),and is equal to 5

Ankit Khokhar
21 Points
12 years ago

for the first question let t=sin x

so e^t - e^(-t) - 4 = 0

so e^2t - 4e^t -1 = 0

so by solving for e^t, we get (2 + 5^(1/2)) or (2 - 5^(1/2))

but e^t is always positive so (2 - 5^(1/2)) is neglected

now   t   belongs to [-1 , 1]

so max. value of e^t is 2.718...

so e^t cannot be (2 + 5^(1/2))

hence no real solution exists.

Now for second, let sin x = t,

so for 1 + 12t - 9t^2  where  t  belongs to [-1 , 1]

as its a(coefficient of t^2)< so parabola will open downward.

find its vertex,

x coordinate=(-b/2a)=(-12/-18)=2/3 so it in the range of t.

Now y coordinate= (-D/4a) (D= discriminant)

so y=(-180/-36)=5

so max value = 5.

Hope you understand.

Abhijat Pandey
31 Points
12 years ago