The number of values of θ in the interval (-(π)/(2), (π)/(2)) such that θ ≠ (n π)/(5) for n=0, ± 1, ± 2 and θ= 5 θ as well as 2 θ= 4 θ is

Correct answer: 3 Given, θ= 5 θ aligned & θ= ((π)/(2)-5 θ) & (π)/(2)-5 θ=n π+θ & 6 θ=(π)/(2)-n π & θ=(π)/(12)-(n π)/(6) & Also, 4 θ= 2 θ= ((π)/(2)-2 θ) & 4 θ=2 n π ±((π)/(2)-2 θ) aligned Taking positive 6 θ=2 n π+(π)/(2) θ=(n π)/(3)+(π)/(12) Taking negative 2 θ=2 n π-(π)/(2) θ=n π-(π)/(4) Above values of θ suggests that there are only 3 common solutions.

JEE Advanced 2010Paper 1

Exam: JEE Advanced
Year: 2010
Paper: Paper 1
Subject: Mathematics
Chapter: Trigonometric Equations

JEE Advanced 2010 · Paper 1 · Mathematics · Trigonometric Equations

The number of values of θ in the interval (-(π)/(2), (π)/(2)) such...

The number of values of

\theta

in the interval

\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

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

Correct answer

Step-by-step solution

Given,

\tan \theta=\cot 5 \theta

\begin{aligned} & \Rightarrow \quad \tan \theta=\tan \left(\frac{\pi}{2}-5 \theta\right) \\ & \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, } \cos 4 \theta=\sin 2 \theta=\cos \left(\frac{\pi}{2}-2 \theta\right) \\ & \Rightarrow \quad 4 \theta=2 n \pi \pm\left(\frac{\pi}{2}-2 \theta\right) \end{aligned}

Taking positive 6 \theta=2 n \pi+\frac{\pi}{2} \Rightarrow \theta=\frac{n \pi}{3}+\frac{\pi}{12} Taking negative 2 \theta=2 n \pi-\frac{\pi}{2} \Rightarrow \theta=n \pi-\frac{\pi}{4} Above values

of \theta

suggests that there are only 3 common solutions.

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