The number of solutions, of the equation esinx

Engineering

Mathematics

General Solution of Trigonometric Equation

Question

The number of solutions, of the equation e^sinx – 2e^–sinx = 2 is ).

more than 2

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Let e^sinx = t

⇒ $t - \frac{2}{t} = 2 \Rightarrow t^{2} - 2 t - 2 = 0$

(t – 1)² = 3

$t - 1 = \pm \sqrt{3}$

$t = 1 \pm \sqrt{3} but t > 0$

$so, e^{sinx} = 1 + \sqrt{3}$

Now, – 1 ≤ sinx ≤ 1

e^–1 ≤ e^sinx ≤ e

e ≅ 2.72 but 1+ $\sqrt{3}$ ≅ 2.73

⇒ No. of solutions = 0