Trigonometrical Equations 3 Question 1
1. For $x \in(0, \pi)$, the equation $\sin x+2 \sin 2 x-\sin 3 x=3$ has
(2014 Adv.)
(a) infinitely many solutions
(b) three solutions
(c) one solution
(d) no solution
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Answer:
Correct Answer: 1. (d)
Solution:
- PLAN For solving this type of questions, obtain the LHS and RHS in equation and examine, the two are equal or not for a given interval.
Given, trigonometrical equation
$ (\sin x-\sin 3 x)+2 \sin 2 x=3 $
$\Rightarrow-2 \cos 2 x \sin x+4 \sin x \cos x=3$
$ \begin{aligned} & {\left[\because \sin C-\sin D=2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}\right. \text { and }} \\ &\sin 2 \theta=2 \sin \theta \cos \theta] \\ & \Rightarrow \quad 2 \sin x(2 \cos x-\cos 2 x)=3 \\ & \Rightarrow \quad 2 \sin x\left(2 \cos x-2 \cos ^{2} x+1\right)=3 \\ & \Rightarrow \quad 2 \sin x \frac{3}{2}-2 \cos x-\frac{1}{2}=3 \\ & \Rightarrow \quad 3 \sin x-3=4 \cos x-\frac{1}{2}^{2} \sin x \end{aligned} $
As $x \in(0, \pi) \quad$ LHS $\leq 0$ and RHS $\geq 0$
For solution to exist, LHS $=$ RHS $=0$
Now, LHS $=0$
$ \begin{aligned} \Rightarrow & & 3 \sin x-3 & =0 \\ \Rightarrow & & \sin x & =1 \\ \Rightarrow & & x & =\frac{\pi}{2} \end{aligned} $
For $\quad x=\frac{\pi}{2}$,
RHS $=4 \quad \cos \frac{\pi}{2}-\frac{1}{2}^{2} \sin \frac{\pi}{2}=4 \quad \frac{1}{4} \quad(1)=1 \neq 0$
$\therefore$ No solution of the equation exists.
