The number of real roots of the equation {e^{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given equation: ${e^{\sin x}} - {e^{-\sin x}} - 4 = 0$.Let ${e^{\sin x}} = y$. Since ${e^{\sin x}} > 0$,we have $y > 0$.The equation becomes $y -

Vedclass · Accepted Answer

None

Vedclass · Answer

(D) Given equation: ${e^{\sin x}} - {e^{-\sin x}} - 4 = 0$.Let ${e^{\sin x}} = y$. Since ${e^{\sin x}} > 0$,we have $y > 0$.The equation becomes $y - \frac{1}{y} - 4 = 0$.Multiplying by $y$,we get ${y^2} - 4y - 1 = 0$.Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $y = \frac{4 \pm \sqrt{16 - 4(1)(-1)}}{2} = \frac{4 \pm \sqrt{20}}{2} = 2 \pm \sqrt{5}$.Since $y > 0$ and $\sqrt{5} \approx 2.236$,we must have $y = 2 + \sqrt{5}$.Thus,${e^{\sin x}} = 2 + \sqrt{5}$.Taking the natural logarithm on both sides: $\sin x = \ln(2 + \sqrt{5})$.Since $\sqrt{5} > 1$,$2 + \sqrt{5} > 3$. Therefore,$\ln(2 + \sqrt{5}) > \ln(3) > 1$.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 number of real roots is $0$ (None).

The number of real roots of the equation ${e^{\sin x}} - {e^{-\sin x}} - 4 = 0$ is:

Similar Questions

If the roots of the equation $Ax^2 + Bx + C = 0$ are $\alpha, \beta$ and the roots of the equation $x^2 + px + q = 0$ are $\alpha^2, \beta^2$,then the value of $p$ will be

The roots of the given equation $2(a^2 + b^2)x^2 + 2(a + b)x + 1 = 0$ are

If $x=\frac{\sqrt{3}+1}{\sqrt{3}-1}$ and $y=\frac{\sqrt{3}-1}{\sqrt{3}+1}$,find the value of $x^{2}+y^{2}$.

If $a$ and $b$ are odd integers,then the roots of the equation $2ax^2 + (2a + b)x + b = 0, a \ne 0,$ will be

Solve the given two equations and select the correct answer from the given options.
$I.$ $(17)^{2} + 144 \div 18 = x$
$II.$ $(26)^{2} - 18 \times 21 = y$

Vedclass Test Series

Exam Paper Generator

Online Exam Module