Question
The number of real roots of the equation esin x−e−sin x−4=0 are
Answer: Option D
:
D
Given equationesinx−e−sinx−4=0
Letesinx=y, then given equation can be written as
y2−4y−1=0 ⇒y=2±√5
But the value of y=esinx is always positive, so
y=2+√5 (∵ 2 < √5
⇒logey=loge(2+√5)⇒ sin x = loge(2+√5)>1
no solution
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:
D
Given equationesinx−e−sinx−4=0
Letesinx=y, then given equation can be written as
y2−4y−1=0 ⇒y=2±√5
But the value of y=esinx is always positive, so
y=2+√5 (∵ 2 < √5
⇒logey=loge(2+√5)⇒ sin x = loge(2+√5)>1
no solution
Was this answer helpful ?
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