For x in (0, pi),the equation sin x + 2 sin 2 | vedclass

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

Question

(D) Given the equation: $\sin x + 2 \sin 2x - \sin 3x = 3$. Using the identity $\sin 3x = 3 \sin x - 4 \sin^3 x$,we substitute: $\sin x + 2(2 \sin x \

Vedclass · Accepted Answer

no solution

Vedclass · Answer

(D) Given the equation: $\sin x + 2 \sin 2x - \sin 3x = 3$. Using the identity $\sin 3x = 3 \sin x - 4 \sin^3 x$,we substitute: $\sin x + 2(2 \sin x \cos x) - (3 \sin x - 4 \sin^3 x) = 3$. $-2 \sin x + 4 \sin x \cos x + 4 \sin^3 x = 3$. Since $\sin x \in (0, 1]$ for $x \in (0, \pi)$,we analyze the maximum value of the expression $f(x) = \sin x + 2 \sin 2x - \sin 3x$. We know that $\sin x \leq 1$,$\sin 2x \leq 1$,and $-\sin 3x \leq 1$. However,these cannot all be $1$ simultaneously for the same $x$. Alternatively,consider the range of $f(x) = \sin x + 4 \sin x \cos x - 3 \sin x + 4 \sin^3 x = 4 \sin x \cos x + 4 \sin^3 x - 2 \sin x$. $f(x) = 2 \sin x (2 \cos x + 2 \sin^2 x - 1) = 2 \sin x (2 \cos x + 2(1 - \cos^2 x) - 1) = 2 \sin x (-2 \cos^2 x + 2 \cos x + 1)$. Let $u = \cos x$,where $u \in (-1, 1)$. The function $g(u) = 2 \sqrt{1-u^2} (-2u^2 + 2u + 1)$. By checking the maximum value,it is found that $f(x) Thus,there is no solution.

For $x \in (0, \pi)$,the equation $\sin x + 2 \sin 2x - \sin 3x = 3$ has

Similar Questions

If the sides of a triangle $a, b, c$ are in $A.P.$,then with usual notations,$a \cos ^2 \frac{C}{2} + c \cos ^2 \frac{A}{2}$ is

In a triangle,if $r_1 = 2r_2 = 3r_3$,then $\frac{a}{b} + \frac{b}{c} + \frac{c}{a}$ is equal to

If $a \cos A = b \cos B$,then $\Delta ABC$ is

In a triangle $ABC,$ ${a^3}\cos (B - C) + {b^3}\cos (C - A) + {c^3}\cos (A - B) = $

In a triangle $ABC$,if $a, b, c$ are in arithmetic progression and the angle $A$ is twice the angle $C$,then $\cos A : \cos B : \cos C =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module