Q. 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 $\sin 2 \theta=\cos 4 \theta,$ is
[JEE 2010, 3]
Q. The positive integer value of n > 3 satisfying the equation
$\frac{1}{\sin \left(\frac{\pi}{\mathrm{n}}\right)}=\frac{1}{\sin \left(\frac{2 \pi}{\mathrm{n}}\right)}+\frac{1}{\sin \left(\frac{3 \pi}{\mathrm{n}}\right)}$ is
[JEE 2011, 4]
Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...
Sol. 7 $\frac{1}{\sin \frac{\pi}{\mathrm{n}}}=\frac{1}{\sin \frac{2 \pi}{\mathrm{n}}}+\frac{1}{\sin \frac{3 \pi}{\mathrm{n}}}$ $\Rightarrow \frac{1}{\sin \frac{\pi}{\mathrm{n}}}-\frac{1}{\sin \frac{3 \pi}{\mathrm{n}}}=\frac{1}{\sin \frac{2 \pi}{\mathrm{n}}}$ $\Rightarrow \frac{\sin \frac{3 \pi}{n}-\sin \frac{\pi}{n}}{\sin \frac{\pi}{n} \sin \frac{3 \pi}{n}}=\frac{1}{\sin \frac{2 \pi}{n}}$ $\Rightarrow \frac{2 \cos \frac{2 \pi}{\mathrm{n}} \sin \frac{\pi}{\mathrm{n}}}{\sin \frac{\pi}{\mathrm{n}} \sin \frac{3 \pi}{\mathrm{n}}}=\frac{1}{\sin \frac{2 \pi}{\mathrm{n}}}$ $\Rightarrow 2 \cos \frac{2 \pi}{\mathrm{n}} \sin \frac{2 \pi}{\mathrm{n}}=\sin \frac{3 \pi}{\mathrm{n}}$ $\Rightarrow \sin \frac{4 \pi}{\mathrm{n}}=\sin \frac{3 \pi}{\mathrm{n}} \Rightarrow \frac{4 \pi}{\mathrm{n}}=\mathrm{K} \pi+(-1)^{\mathrm{K}} \frac{3 \pi}{\mathrm{n}}$ If $\mathrm{K}=2 \mathrm{m} \quad \Rightarrow \quad \frac{\pi}{\mathrm{n}}=2 \mathrm{m} \pi$ $\Rightarrow \quad n=\frac{1}{2 m} \quad \Rightarrow n=\frac{1}{2}, \frac{1}{4}, \frac{1}{6} \ldots \ldots$ If $\mathrm{K}=2 \mathrm{m}+1 \Rightarrow \frac{7 \pi}{\mathrm{n}}=(2 \mathrm{m}+1) \pi$ $\Rightarrow \mathrm{n}=\frac{7}{2 \mathrm{m}+1} \quad \Rightarrow \quad \mathrm{n}=7, \frac{7}{3}, \frac{7}{5} \ldots \ldots$ Possible value of n is 7
Q. Let $\theta, \varphi \in[0,2 \pi]$ be such that $2 \cos \theta(1-\sin \varphi)=\sin ^{2} \theta\left(\tan \frac{\theta}{2}+\cot \frac{\theta}{2}\right) \cos \varphi-1, \tan (2 \pi-\theta)>0$
and $-1<\sin \theta<-\frac{\sqrt{3}}{2} .$ Then $\varphi$ cannot satisfy-
(A) $0<\varphi<\frac{\pi}{2}$
(B) $\frac{\pi}{2}<\varphi<\frac{4 \pi}{3}$
(C) $\frac{4 \pi}{3}<\varphi<\frac{3 \pi}{2}$
(D) $\frac{3 \pi}{2}<\varphi<2 \pi$
[JEE 2012, 4M]
Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...
Sol. (A,C,D)
Q. For $\mathrm{x} \in(0, \pi),$ the equation $\sin \mathrm{x}+2 \sin 2 \mathrm{x}-\sin 3 \mathrm{x}=3 \mathrm{has}$
(A) infinitely many solutions
(B) three solutions
(C) one solution
(D) no solution
[JEE(Advanced)-2014, 3(–1)]
Q. The number of distinct solutions of equation $\frac{5}{4} \cos ^{2} 2 x+\cos ^{4} x+\sin ^{4} x+\cos ^{6} x+\sin ^{6} x=2$ in the interval $[0,2 \pi]$ is
[JEE 2015, 4M, –0M]
Q. Let $S=\left\{x \in(-\pi, \pi): x \neq 0, \pm \frac{\pi}{2}\right\} .$ The sum of all distinct solution of the equation $\sqrt{3} \sec x+\csc x+2(\tan x-\cot x)=0$ in the set $S$ is equal to $-$
(A) $-\frac{7 \pi}{9}$
(B) $-\frac{2 \pi}{9}$
(C) 0
(D) $\frac{5 \pi}{9}$
[JEE(Advanced)-2016]
Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...
Sol. (C) $\sqrt{3} \sin x+\cos x=2 \cos 2 x$ $\Rightarrow \cos 2 x=\cos \left(x-\frac{\pi}{3}\right)$ $\Rightarrow 2 x=2 n \pi \pm\left(x-\frac{\pi}{3}\right)$ $\quad \quad x=(6 n-1) \frac{\pi}{3}$ or $(6 n+1) \frac{\pi}{9}$ $\Rightarrow x=-\frac{\pi}{3}, \frac{\pi}{9}, \frac{7 \pi}{9}$ and $-\frac{5 \pi}{9}$ in $(-\pi, \pi)$ $\Rightarrow \operatorname{sum}=0$
Q. Let $a, b, c$ be three non-zero real numbers such that the equation $\sqrt{3} a \cos x+2 b \sin x=c, \quad x$ $\in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ has two distinct real roots $\alpha$ and $\beta$ with $\alpha+\beta=\frac{\pi}{3} .$ Then the value of $\frac{b}{a}$ is $-$
$=$
[JEE(Advanced)-2018]
Op
Awesome…but can be solved in da methods too
Good beta. Keep studying
Awesome…but can be solved in different methods too
Wht my dadd?
Nice question but those can be solved by further methods so not harder
Great ! Thanks for it .
LIKE A BAWWS!!!
Problems are tough but thanks for providing these questions for practise🙏🙏🙏
Nice questions thank you
lol
👌❣️👌
Easy
good