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 $\operatorname{sum}=\frac{13 \pi}{9}$
For printable Trigonometric Equations practice, the JEE PYQ chapter wise page provides an organised route to other Maths topics.
Recognising which identity to apply is half the battle here — the trigo class 11 formula page is the fastest way to revise before this set.
If full mock-exam conditions suit you better than isolated Trigonometric Equations questions, our JEE Main previous year paper is the natural next step.
