Trigonometrical Equations Ques 45
- Find the coordinates of the points of intersection of the curves $y=\cos x, y=\sin 3 x$, if $-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$.
(1982, 3M)
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Answer:
Correct Answer: 45.$\frac{\pi}{8}, \cos \frac{\pi}{8} \quad \frac{\pi}{4}, \cos \frac{\pi}{4}-\frac{3 \pi}{8}, \cos \frac{3 \pi}{8}$
Solution:
Formula:
- The point of intersection is given by
$ \begin{aligned} \sin 3 x & =\cos x=\sin (\frac{\pi}{2}-x) \\ \Rightarrow \quad 3 x & =n \pi+(-1)^{n} (\frac{\pi}{2}-x) \end{aligned} $
(i) Let $n$ be even i.e. $n=2 m$
$ \begin{aligned} \Rightarrow & & 3 x & =2 m \pi+\frac{\pi}{2}-x \\ \Rightarrow & & n & =\frac{m \pi}{2}+\frac{\pi}{8}……(i) \end{aligned} $
(ii) Let $n$ be odd i.e. $n=(2 m+1)$
$ \begin{array}{ll} \therefore & 3 x=(2 m+1) \pi-(\frac{\pi}{2}-x) \\ \Rightarrow & 3 x=2 m \pi+\frac{\pi}{2}+x \\ \Rightarrow & x=m \pi+\frac{\pi}{4} ……(ii) \\ \text { Now, } & -\frac{\pi}{2} \leq x \leq \frac{\pi}{2} \\ \Rightarrow & x=\frac{\pi}{8}, \frac{\pi}{4}-\frac{3 \pi}{8} \quad \text { [from Eqs. (i) and (ii)] } \end{array} $
Thus, points of intersection are
$ (\frac{\pi}{8}, \cos \frac{\pi}{8}) \quad (\frac{\pi}{4}, \cos \frac{\pi}{4}) \quad (-\frac{3 \pi}{8}, \cos \frac{3 \pi}{8}) $
BETA
