The equation e^{sin x} - e^{sin(-x)} - 4 = 0 | vedclass

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(B) Given equation is $e^{\sin x} - e^{-\sin x} - 4 = 0$. Let $e^{\sin x} = t$. Since $e^{\sin x} > 0$,we must have $t > 0$. The equation becomes $t -

Vedclass · Accepted Answer

no real roots

Vedclass · Answer

(B) Given equation is $e^{\sin x} - e^{-\sin x} - 4 = 0$. Let $e^{\sin x} = t$. Since $e^{\sin x} > 0$,we must have $t > 0$. The equation becomes $t - \frac{1}{t} - 4 = 0$. Multiplying by $t$,we get $t^2 - 4t - 1 = 0$. Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $t = \frac{4 \pm \sqrt{16 + 4}}{2} = \frac{4 \pm 2\sqrt{5}}{2} = 2 \pm \sqrt{5}$. Since $t > 0$,we reject $t = 2 - \sqrt{5}$ (as $2 - \sqrt{5} Thus,$e^{\sin x} = 2 + \sqrt{5}$. Taking the natural logarithm on both sides,$\sin x = \ln(2 + \sqrt{5})$. Since $\sqrt{5} \approx 2.236$,$2 + \sqrt{5} \approx 4.236$. We know that $\ln(e) = 1$ and $e \approx 2.718$. Since $4.236 > 2.718$,$\ln(2 + \sqrt{5}) > 1$. However,the range of $\sin x$ is $[-1, 1]$. Since $\ln(2 + \sqrt{5}) > 1$,there is no real value of $x$ that satisfies this equation. Therefore,the equation has no real roots.

The equation $e^{\sin x} - e^{\sin(-x)} - 4 = 0$ has

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