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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........????
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
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.
PLZ someone answer this.....
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