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Show that the equation e- sin x –e - sin x – 4 = 0 has no real solution.

Show that the equation e-sin x –e- sin x – 4 = 0 has no real solution.

Grade:upto college level

2 Answers

Navjyot Kalra
askIITians Faculty 654 Points
7 years ago
Sol. esin x – e –sin x – 4 = 0
Let esin x = y then e-sin x = 1/y
∴ Equation becomes, y – 1/y – 4 = 0
⇒ y2 – 4y – 1 = 0
⇒ y = 2 + √5,2 - √5
But y is real +ve number,
∴ y ≠ 2 - √5 ⇒ y = 2 + √5
⇒ esin x = 2 + √5 ⇒ sin x = loge (2 + √5)
But 2 + √5 > e ⇒ loge (2 + √5) > logee
⇒ loge (2 + √5) > 1 Hence, sin x > 1
Which is not possible.
∴ Given equation has no real solution.
ankit singh
askIITians Faculty 614 Points
one year ago

esinx – e-sinx – 4 = 0

t = esinx

t – 1/t = 4

t2 – 4t – 1 = 0

t = 4 ± √16 + 4 / (2)

t = 4 ± 2√5 / (2)

t = 2 ± √5

esinx = 2 ± √5

-1 ≤ sinx ≤ 1

1/e ≤ esinx ≤ e

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