Trigonometrical Equations Ques 29
- If $0 \leq x<\frac{\pi}{2}$, then the number of values of $x$ for which $\sin x-\sin 2 x+\sin 3 x=0$, is
(a) 2
(b) 3
(c) 1
(d) 4
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Answer:
Correct Answer: 29.(a)
Solution:
Formula:
- We have, $\sin x-\sin 2 x+\sin 3 x=0$
$\Rightarrow(\sin x+\sin 3 x)-\sin 2 x=0$
$\Rightarrow 2 \sin (\frac{x+3 x}{2}) \cos (\frac{x-3 x}{2})-\sin 2 x=0$
$ \left[\because \sin C+\sin D=2 \sin (\frac{C+D}{2}) \cos (\frac{C-D}{2})\right] $
$ \begin{array}{lrl} \Rightarrow & 2 \sin 2 x \cos x-\sin 2 x=0 & {[\because \cos (-\theta)=\cos \theta]} \\ \Rightarrow & \sin 2 x(2 \cos x-1)=0 & \\ \Rightarrow & \sin 2 x=0 \text { or } 2 \cos x-1=0 \\ \Rightarrow & 2 x=0, \pi, \ldots \text { or } \cos x=\frac{1}{2} \\ \Rightarrow & x=0, \frac{\pi}{2} \ldots \text { or } x=\frac{\pi}{3} \end{array} $
In the interval $[0, \frac{\pi}{2})$ only two values satisfy, namely $x=0$ and $x=\frac{\pi}{3}$.
BETA
