JEE Main Trigonometric Equations Practice Questions With Solutions

Let $|\cos \theta \cos (60-\theta )\cos (60+\theta )|\le \frac{1}{8},\theta ϵ[0,2\pi ]$\left|\right. cos ⁡ \theta cos ⁡ \left(\right. 60 - \theta \left.\right) cos ⁡ \left(\right. 60 + \theta \left.\right) \left|\right. \leq \frac{1}{8} , \theta \epsilon \left[\right. 0 , 2 \pi \left]\right.. Then, the sum of all $\theta \in [0,2\pi ]$\theta \in \left[\right. 0 , 2 \pi \left]\right., where $\cos 3\theta$cos ⁡ 3 \theta attains its maximum value, is :

If $2{\sin }^{3}x+\sin 2x\cos x+4\sin x-4=0$2 sin^{3} ⁡ x + sin ⁡ 2 x cos ⁡ x + 4 sin ⁡ x - 4 = 0 has exactly 3 solutions in the interval $[0,\frac{\mathrm{n}\pi }{2}],\mathrm{n}\in \mathrm{N}$\left[\right. 0 , \frac{n \pi}{2} \left]\right. , n \in N, then the roots of the equation $x^2+\mathrm{n}x+(\mathrm{n}-3)=0$x^{2} + n x + \left(\right. n - 3 \left.\right) = 0 belong to :

The sum of the solutions $x\in \mathbb{R}$x \in \mathbb{R} of the equation $\frac{3\cos 2x+{\cos }^{3}2x}{{\cos }^{6}x-{\sin }^{6}x}=x^3-x^2+6$\frac{3 cos ⁡ 2 x + cos^{3} ⁡ 2 x}{cos^{6} ⁡ x - sin^{6} ⁡ x} = x^{3} - x^{2} + 6 is

If $\alpha ,-\frac{\pi }{2}<\alpha <\frac{\pi }{2}$\alpha , - \frac{\pi}{2} < \alpha < \frac{\pi}{2} is the solution of $4\cos \theta +5\sin \theta =1$4 cos ⁡ \theta + 5 sin ⁡ \theta = 1, then the value of $\tan \alpha$tan ⁡ \alpha is

If $2{\tan }^2\theta -5\sec \theta =1$2 tan^{2} ⁡ \theta - 5 sec ⁡ \theta = 1 has exactly 7 solutions in the interval $[0,\frac{n\pi }{2}]$\left[\right. 0 , \frac{n \pi}{2} \left]\right., for the least value of $n\in \mathbf{N}$n \in \mathbf{N}, then $\sum _{k=1}^n\frac{k}{2^k}$\sum_{k = 1}^{n} \frac{k}{2^{k}} is equal to:

The number of elements in the set

$S=\{\theta \in [0,2\pi ]:3{\cos }^4\theta -5{\cos }^2\theta -2{\sin }^6\theta +2=0\}$S = \left{\right. \theta \in \left[\right. 0 , 2 \pi \left]\right. : 3 cos^{4} ⁡ \theta - 5 cos^{2} ⁡ \theta - 2 sin^{6} ⁡ \theta + 2 = 0 \left.\right} is :

Let $S=\{x\in (-\frac{\pi }{2},\frac{\pi }{2}):9^{1-{\tan }^{2}x}+9^{{\tan }^{2}x}=10\}$S = \left{\right. x \in \left(\right. - \frac{\pi}{2} , \frac{\pi}{2} \left.\right) : 9^{1 - tan^{2} ⁡ x} + 9^{tan^{2} ⁡ x} = 10 \left.\right} and$\beta =\sum _{x\in S}{\tan }^2\left(\frac{x}{3}\right)$\beta = \underset{x \in S}{\sum} tan^{2} ⁡ \left(\right. \frac{x}{3} \left.\right), then $\frac{1}{6}(\beta -14{)}^2$\frac{1}{6} \left(\right. \beta - 14 \left.\right)^{2} is equal to :

The number of elements in the set $S=\{x\in \mathbb{R}:2\cos \left(\frac{x^2+x}{6}\right)=4^x+4^{-x}\}$S = \left{\right. x \in \mathbb{R} : 2 cos ⁡ \left(\right. \frac{x^{2} + x}{6} \left.\right) = 4^{x} + 4^{- x} \left.\right} is :

Let $S=\{\theta \in (0,\frac{\pi }{2}):\sum _{m=1}^9\sec (\theta +(m-1)\frac{\pi }{6})\sec (\theta +\frac{m\pi }{6})=-\frac{8}{\sqrt{3}}\}$S = \left{\right. \theta \in \left(\right. 0 , \frac{\pi}{2} \left.\right) : \sum_{m = 1}^{9} sec ⁡ \left(\right. \theta + \left(\right. m - 1 \left.\right) \frac{\pi}{6} \left.\right) sec ⁡ \left(\right. \theta + \frac{m \pi}{6} \left.\right) = - \frac{8}{\sqrt{3}} \left.\right}. Then