SATHEE: Trigonometrical Equations 3 Question 17

Trigonometrical Equations 3 Question 17

18. Show that the equation $e^{\sin x} - e^{- \sin x} - 4 = 0$ has no real solution.

$(1982, 2 M)$

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Solution:

Given,

$e^{\sin x} - \frac{1}{e^{\sin x}} = 4$

$\begin{aligned} \Rightarrow {(e^{\sin x})}^{2} - 4 (e^{\sin x}) - 1 = 0 \\ \Rightarrow e^{\sin x} = \frac{4 \pm \sqrt{16 + 4}}{2} = 2 \pm \sqrt{5} \end{aligned}$

But since, $e \sim 2.72$ and we know, $0 < e^{\sin x} < e$

$∴ e^{\sin x} = 2 \pm \sqrt{5}$ is not possible.

Hence, it does not exist any solution.

Trigonometrical Equations 3 Question 17

18. Show that the equation $e^{\sin x} - e^{- \sin x} - 4 = 0$ has no real solution.

Solution:

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Trigonometrical Equations 3 Question 17

18. Show that the equation esin⁡x−e−sin⁡x−4=0 has no real solution.

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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18. Show that the equation $e^{\sin x} - e^{- \sin x} - 4 = 0$ has no real solution.