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.SimulatorPrevious Years AIEEE/JEE Main Questions
Q. Let A and B denote the statements$\mathbf{A}: \cos \alpha+\cos \beta+\cos \gamma=0$$\mathbf{B}: \sin \alpha+\sin \beta+\sin \gamma=0If \cos (\beta-\gamma)+\cos (\gamma-\alpha)+\cos (\alpha-\beta)=-\frac{3}{2}, then :-(1) Both A and B are true(2) Both A and B are false(3) A is true and B is false(4) A is false and B is true [AIEEE 2009] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (1)\cos (\beta-\gamma)+\cos (\gamma-\alpha)+\cos (\alpha-\beta)=-\frac{3}{2}$$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$$(\sin \alpha+\sin \beta+\sin \gamma)^{2}+(\cos \alpha+\cos \beta+\cos \gamma)^{2}=0Only Possible when\sin \alpha+\sin \beta+\sin \gamma=0$$\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]

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Sol. (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}$$\Rightarrow 3 \theta=\frac{2 \pi}{3}, \frac{4 \pi}{3}, \frac{8 \pi}{3}, \frac{10 \pi}{3} \quad \Rightarrow \quad \theta=\frac{2 \pi}{9}, \frac{4 \pi}{9}, \frac{8 \pi}{9}, \frac{10 \pi}{9}but \frac{10 \pi}{9} \notin(0, \pi)So, \theta=\frac{\pi}{4}, \frac{\pi}{2}, \frac{3 \pi}{4}, \frac{2 \pi}{9}, \frac{4 \pi}{9}, \frac{8 \pi}{9} Q. If 0 \leq x<2 \pi, then the number of real values of x, which satisfy the equation\cos x+\cos 2 x+\cos 3 x+\cos 4 x=0, is : –(1) 9 (2) 3 (3) 5 (4) 7 [JEE Mains 2016] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (4)2 \cos 2 x \cos x+2 \cos 3 x \cos x=0$$\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]

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Sol. (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$\operatorname{sum}=\frac{13 \pi}{9}$

• October 4, 2021 at 7:47 am

even each attempt contains more triogo questions then this

• April 15, 2021 at 2:03 pm

pure 15 saal me 4 questions hi aae hai kya jee main bc?

• February 19, 2021 at 8:03 pm

Aur questions nahi they ya fer jaga nahi tha daal ne ke liye?

• September 7, 2021 at 6:54 pm

🤣🤣🤣🤣

• January 2, 2021 at 11:03 am

provide more questions

• October 14, 2020 at 2:39 pm

Very less Qns

• October 21, 2020 at 3:46 pm

Good questions but add more question from jee main 2017-2020 papers

• October 8, 2020 at 11:52 am

You could have added more questions……

• September 23, 2020 at 5:59 pm

Very less questions

• September 3, 2020 at 12:04 pm

Good

• August 27, 2020 at 8:41 pm

Kishan Ka Sala Hun

• August 27, 2020 at 11:43 am

Thank you.

• August 15, 2020 at 5:44 pm

This app is so worst that it is not useful for anyone because
1.solutions are not clear
2.size of letters are very small

• May 29, 2020 at 8:38 am

Thanks sir

• May 13, 2020 at 1:32 pm

Good app ! Super I like this app the most

• March 3, 2020 at 5:41 pm

2019 2020 ke jameen ke previous year paper with solution Dal do please