Trigonometrical Equations Ques 39
- The number of values of $\theta$ in the interval $-\frac{\pi}{2}, \frac{\pi}{2}$ such that $\theta \neq \frac{n \pi}{5}$ for $n=0, \pm 1, \pm 2$ and $\tan \theta=\cot 5 \theta$ as well as $\sin 2 \theta=\cos 4 \theta$ is……
(2010)
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Answer:
Correct Answer: 39.3
Solution:
Formula:
- Given, $\tan \theta=\cot 5 \theta$
$ \begin{alignedat} & \Rightarrow \quad \tan \theta=\tan (\frac{\pi}{2}-5 \theta) \\ & \Rightarrow \quad \frac{\pi}{2}-5 \theta=n \pi+\theta \\ & \Rightarrow \quad 6 \theta=\frac{\pi}{2}+n \pi \\ & \Rightarrow \quad \theta=\frac{\pi}{12}-\frac{n \pi}{6} \\ & \text { Also, } \quad \cos 4 \theta=\sin 2 \theta=\cos (\frac{\pi}{2}-2 \theta) \\ & \Rightarrow \quad 4 \theta=2 n \pi \pm (\frac{\pi}{2}-2 \theta) \end{aligned} $
Taking positive sign,
$ \begin{alignedat} 6 \theta & =2 n \pi+\frac{\pi}{2} \\ \Rightarrow \quad \theta & =\frac{n \pi}{3}+\frac{\pi}{12} \end{aligned} $
Taking the negative sign,
$ 2 \theta=2 n \pi-\frac{\pi}{2} \Rightarrow \theta=n \pi-\frac{\pi}{4} $
Above values of $\theta$ suggest that there are only 3 distinct solutions.
BETA
