Exams Joint Entrance Examination Mathematics Trigonometric Ratios and Identities Question 19 Question 19 of 298 Medium MCQ
The number of solutions, of the equation \[e^{\sin x}-2 e^{-\sin x}=2\], is :
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Take \[e^{\sin x}=t(t>0)\]
\[ ⇒ t − 2 t = 2 ⇒ t 2 − 2 t = 2 ⇒ t 2 − 2 t − 2 = 0 ⇒ t 2 − 2 t + 1 = 3 ⇒ ( t − 1 ) 2 = 3 ⇒ t = 1 ± 3 ⇒ t = 1 ± 1.73 ⇒ t = 2.73 or − 0.73 (rejected as t > 0 ) ⇒ e sin x = 2.73 ⇒ log e e sin x = log e 2.73 ⇒ sin x = log e 2.73 > 1 \begin{aligned}
& \Rightarrow \mathrm{t}-\frac{2}{\mathrm{t}}=2 \\
& \Rightarrow \frac{\mathrm{t}^2-2}{\mathrm{t}}=2 \\
& \Rightarrow \mathrm{t}^2-2 \mathrm{t}-2=0 \\
& \Rightarrow \mathrm{t}^2-2 \mathrm{t}+1=3 \\
& \Rightarrow(\mathrm{t}-1)^2=3 \\
& \Rightarrow \mathrm{t}=1 \pm \sqrt{3} \\
& \Rightarrow \mathrm{t}=1 \pm 1.73 \\
& \Rightarrow \mathrm{t}=2.73 \text { or }-0.73 \text { (rejected as } \mathrm{t}>0) \\
& \Rightarrow \mathrm{e}^{\sin \mathrm{x}}=2.73 \\
& \Rightarrow \log _{\mathrm{e}} \mathrm{e}^{\sin \mathrm{x}}=\log _{\mathrm{e}} 2.73 \\
& \Rightarrow \sin \mathrm{x}=\log _{\mathrm{e}} 2.73>1
\end{aligned} ⇒ t − t 2 = 2 ⇒ t t 2 − 2 = 2 ⇒ t 2 − 2 t − 2 = 0 ⇒ t 2 − 2 t + 1 = 3 ⇒ ( t − 1 ) 2 = 3 ⇒ t = 1 ± 3 ⇒ t = 1 ± 1.73 ⇒ t = 2.73 or − 0.73 (rejected as t > 0 ) ⇒ e s i n x = 2.73 ⇒ log e e s i n x = log e 2.73 ⇒ sin x = log e 2.73 > 1 \]
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