Trigonometric Equation - JEE Advanced Previous Year Questions with Solutions

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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]
Ans. 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]
Ans. 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]
Ans. (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)]
Ans. (D)
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]
Ans. 8
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]
Ans. (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]
Ans. 0.5

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Comments

Mridul Agarwaal
Oct. 15, 2022, 2:55 p.m.
Good Questions!
Siya
Jan. 16, 2022, 1:36 p.m.
Good question 👍👍 They can also be solve by different methods.🤔
Arjun pratap Singh
July 3, 2021, 6:42 p.m.
Op
Shibz
April 9, 2021, 9:38 p.m.
Awesome...but can be solved in da methods too
Mihir bessanio ( shibi bessanios daddy)
March 5, 2021, 5:03 p.m.
Good beta. Keep studying
Shibi bessanio
Feb. 18, 2021, 8:39 p.m.
Nice question but those can be solved by further methods so not harder
Shruti Wadatkar
Nov. 28, 2020, 9:07 a.m.
Great ! Thanks for it .
JAMES S MURRAY
Nov. 5, 2020, 8:48 p.m.
LIKE A BAWWS!!!
Reshma
Sept. 24, 2020, 9:07 a.m.
Problems are tough but thanks for providing these questions for practise🙏🙏🙏
Harshmohan
Aug. 11, 2020, 10:56 a.m.
Nice questions thank you
Israel
June 27, 2020, 11:20 a.m.
👌❣️👌
boss
April 20, 2020, 11:33 a.m.
good
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