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Trigonometric Equation - JEE Main Previous Year Question with Solutions

JEE Main Previous Year Question of Math with Solutions are available at eSaral. Practicing JEE Main Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas. eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects. Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more. Download eSaral app for free study material and video tutorials. Simulator Previous Years AIEEE/JEE Main Questions
Q. Let A and B denote the statements $\mathbf{A}: \cos \alpha+\cos \beta+\cos \gamma=0$ $\mathbf{B}: \sin \alpha+\sin \beta+\sin \gamma=0$ If $\cos (\beta-\gamma)+\cos (\gamma-\alpha)+\cos (\alpha-\beta)=-\frac{3}{2},$ then $:-$ (1) Both A and B are true (2) Both A and B are false (3) A is true and B is false (4) A is false and B is true [AIEEE 2009]
Ans. (1) $\cos (\beta-\gamma)+\cos (\gamma-\alpha)+\cos (\alpha-\beta)=-\frac{3}{2}$ $2 \cos (\beta-\gamma)+2 \cos (\gamma-\alpha)+2 \cos (\alpha-\beta)=-3$ $1+1+1+2(\cos \beta \cos \gamma+\sin \beta \sin \gamma)+2(\cos \gamma \cos \alpha+\sin \gamma \sin \alpha)+2(\cos \alpha \cos \beta+\sin \alpha \sin \beta)$ $=0$ $\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)+\left(\sin ^{2} \beta+\cos ^{2} \beta\right)+\left(\sin ^{2} \gamma+\cos ^{2} \gamma\right)+2 \cos \alpha \cos \beta+2 \cos \beta \cos \gamma+2 \cos$ $\gamma \cos \alpha$ $+2 \sin \alpha \sin \beta+2 \sin \beta \sin \gamma+2 \sin \gamma \sin \alpha=0$ $\left(\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma+2 \sin \alpha \sin \beta+2 \sin \beta \sin \gamma\right.$ $+2 \sin \gamma \sin \alpha)+\left(\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma\right.$ $+2 \cos \alpha \cos \beta+\cos \beta \cos \gamma+\cos \gamma \cos \alpha)=0$ $(\sin \alpha+\sin \beta+\sin \gamma)^{2}+(\cos \alpha+\cos \beta+\cos \gamma)^{2}=0$ Only Possible when $\sin \alpha+\sin \beta+\sin \gamma=0$ $\cos \alpha+\cos \beta+\cos \gamma=0$
Q. The possible values of $\theta \in(0, \pi)$ such that $\sin (\theta)+\sin (4 \theta)+\sin (7 \theta)=0$ are: (1) $\frac{2 \pi}{9}, \frac{\pi}{4}, \frac{4 \pi}{9}, \frac{\pi}{2}, \frac{3 \pi}{4}, \frac{8 \pi}{9}$ (2) $\frac{\pi}{4}, \frac{5 \pi}{12}, \frac{\pi}{2}, \frac{2 \pi}{3}, \frac{3 \pi}{4}, \frac{8 \pi}{9}$ (3) $\frac{2 \pi}{9}, \frac{\pi}{4}, \frac{\pi}{2}, \frac{2 \pi}{3}, \frac{3 \pi}{4}, \frac{35 \pi}{36}$ (4) $\frac{2 \pi}{9}, \frac{\pi}{4}, \frac{\pi}{2}, \frac{2 \pi}{3}, \frac{3 \pi}{4}, \frac{8 \pi}{9}$ [AIEEE 2011]
Ans. (1) $\sin \theta+\sin 4 \theta+\sin 7 \theta=0$ $2 \sin \left(\frac{\theta+7 \theta}{2}\right) \cos \left(\frac{7 \theta-\theta}{2}\right)+\sin 4 \theta=0$ $\Rightarrow \sin 4 \theta[2 \cos 3 \theta+1]=0$ $\Rightarrow \sin 4 \theta=0 \Rightarrow 4 \theta=0, \pi, 2 \pi, 3 \pi, 4 \pi$ $\Rightarrow \theta=0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3 \pi}{4}, \pi$ but 0 and $\pi$ are not included. and $2 \cos 3 \theta+1=0 \Rightarrow \cos 3 \theta=\frac{-1}{2}$ $\Rightarrow 3 \theta=\frac{2 \pi}{3}, \frac{4 \pi}{3}, \frac{8 \pi}{3}, \frac{10 \pi}{3} \quad \Rightarrow \quad \theta=\frac{2 \pi}{9}, \frac{4 \pi}{9}, \frac{8 \pi}{9}, \frac{10 \pi}{9}$ but $\frac{10 \pi}{9} \notin(0, \pi)$ So, $\theta=\frac{\pi}{4}, \frac{\pi}{2}, \frac{3 \pi}{4}, \frac{2 \pi}{9}, \frac{4 \pi}{9}, \frac{8 \pi}{9}$
Q. If $0 \leq x<2 \pi,$ then the number of real values of $x,$ which satisfy the equation $\cos x+\cos 2 x+\cos 3 x+\cos 4 x=0,$ is : - (1) 9        (2) 3           (3) 5          (4) 7 [JEE Mains 2016]
Ans. (4) $2 \cos 2 x \cos x+2 \cos 3 x \cos x=0$ $\Rightarrow 2 \cos x(\cos 2 x+\cos 3 x)=0$ $2 \cos x 2 \cos 5 x / 2 \cos x / 2=0$ $x=\frac{\pi}{2}, \frac{3 \pi}{2}, \pi, \frac{\pi}{5}, \frac{3 \pi}{5}, \frac{7 \pi}{5}, \frac{9 \pi}{5}$ 7 Solutions
Q. If sum of all the solutions of the equation $8 \cos x \cdot\left(\cos \left(\frac{\pi}{6}+x\right) \cdot \cos \left(\frac{\pi}{6}-x\right)-\frac{1}{2}\right)=1$ in $[0, \pi]$ is $k \pi,$ then $k$ is equal to : (1) $\frac{13}{9}$ (2) $\frac{8}{9}$ (3) $\frac{20}{9}$ ( 4)$\frac{2}{3}$ [JEE Mains 2016]
Ans. (1) $8 \cos x\left(\cos ^{2} \frac{\pi}{6}-\sin ^{2} x-\frac{1}{2}\right)=1$ $\Rightarrow 8 \cos x\left(\frac{1}{4}-\left(1-\cos ^{2} x\right)\right)=1$ $\Rightarrow 8 \cos x\left(\cos ^{2} x-\frac{3}{4}\right)=1$ $\Rightarrow 2 \cos 3 x=1 \Rightarrow \cos 3 x=\frac{1}{2}$ $\therefore 3 x+2 n \pi \pm \frac{\pi}{3}, n \in I$ $\Rightarrow \mathrm{x}=\frac{2 \mathrm{n} \pi}{3} \pm \frac{\pi}{9}$ $\ln \mathrm{x} \in[0, \pi]: \mathrm{x}=\frac{\pi}{9}, \frac{2 \pi}{3}+\frac{\pi}{9}, \frac{2 \pi}{3}-\frac{\pi}{9}$ only $\operatorname{sum}=\frac{13 \pi}{9}$

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Rajdeep Sarkar
Oct. 4, 2021, 7:47 a.m.
even each attempt contains more triogo questions then this
Namo kaul from unacademy
April 15, 2021, 2:03 p.m.
pure 15 saal me 4 questions hi aae hai kya jee main bc?
Real Namo Call
May 31, 2023, 6:35 a.m.
भाग यहासे फेक ऐस्स नमो कौल, बिच एस्स निग्गा
Random Guy
Feb. 19, 2021, 8:03 p.m.
Aur questions nahi they ya fer jaga nahi tha daal ne ke liye?
Jan. 2, 2021, 11:03 a.m.
provide more questions
Oct. 14, 2020, 2:39 p.m.
Very less Qns Plss add more
Oct. 8, 2020, 11:52 a.m.
You could have added more questions......
Sept. 23, 2020, 5:59 p.m.
Very less questions
Sept. 3, 2020, 12:04 p.m.
Aug. 27, 2020, 8:41 p.m.
Kishan Ka Sala Hun
Aug. 27, 2020, 11:43 a.m.
Thank you.
B M Gowtham
Aug. 15, 2020, 5:44 p.m.
This app is so worst that it is not useful for anyone because are not clear 2.size of letters are very small
May 29, 2020, 8:38 a.m.
Thanks sir
Nikhil varanasy
May 13, 2020, 1:32 p.m.
Good app ! Super I like this app the most
Rahul Rathore
March 3, 2020, 5:41 p.m.
2019 2020 ke jameen ke previous year paper with solution Dal do please

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