Trigonometric Equations JEE Main previous year questions with solutions

5 solved JEE Main questions on Trigonometric Equations, free to read — no sign-in needed. The full chapter has 57 questions; sign in to attempt the remaining 52 in the exam simulator.

1. Q1JEE Main 2026 (06 Apr, Shift 1)Solving Trigonometric Equation
   Let $S = \{\theta \in (-2\pi, 2\pi) : \cos\theta + 1 = \sqrt{3}\sin\theta\}$. Then $\sum_{\theta \in S}\theta$ is equal to:
   1. A.$-\dfrac{2\pi}{3}$
   2. B.$-\dfrac{4\pi}{3}$
   3. C.$\dfrac{2\pi}{3}$
   4. D.$\dfrac{4\pi}{3}$
   Show answer & solution

   Answer: (B)

   Given equation: $\cos\theta + 1 = \sqrt{3}\sin\theta$ Using half-angle formulas, we get: $2\cos^2\dfrac{\theta}{2} = 2\sqrt{3}\sin\dfrac{\theta}{2}\cos\dfrac{\theta}{2}$ $2\cos\dfrac{\theta}{2} \left( \cos\dfrac{\theta}{2} - \sqrt{3}\sin\dfrac{\theta}{2} \right) = 0$ This gives two cases: Case 1: $\cos\dfrac{\theta}{2} = 0$ $\Rightarrow \dfrac{\theta}{2} = (2k+1)\dfrac{\pi}{2} \Rightarrow \theta = (2k+1)\pi$ Since $\theta \in (-2\pi, 2\pi)$, the possible values are $\theta = -\pi, \pi$. Case 2: $\cos\dfrac{\theta}{2} - \sqrt{3}\sin\dfrac{\theta}{2} = 0$ $\Rightarrow \tan\dfrac{\theta}{2} = \dfrac{1}{\sqrt{3}}$ $\Rightarrow \dfrac{\theta}{2} = k\pi + \dfrac{\pi}{6} \Rightarrow \theta = 2k\pi + \dfrac{\pi}{3}$ Since $\theta \in (-2\pi, 2\pi)$, the possible values are: For $k = 0$, $\theta = \dfrac{\pi}{3}$ For $k = -1$, $\theta = -2\pi + \dfrac{\pi}{3} = -\dfrac{5\pi}{3}$ Thus, the set of solutions is $S = \left\{-\pi, \pi, \dfrac{\pi}{3}, -\dfrac{5\pi}{3}\right\}$. Sum of all solutions in $S$: $\sum_{\theta \in S}\theta = -\pi + \pi + \dfrac{\pi}{3} - \dfrac{5\pi}{3} = -\dfrac{4\pi}{3}$ Answer: $-\dfrac{4\pi}{3}$
2. Q2JEE Main 2011Solving Trigonometric Inequalities
   If $A=\sin ^2 x+\cos ^4 x$, then for all real $x$
   1. A.$\frac{13}{16} \leq \mathrm{A} \leq 1$
   2. B.$1 \leq \mathrm{A} \leq 2$
   3. C.$\frac{3}{4} \leq \mathrm{A} \leq \frac{13}{16}$
   4. D.$\frac{3}{4} \leq \mathrm{A} \leq 1$
   Show answer & solution

   Answer: (D)

   $A=\sin ^2 x+\cos ^4 x=\frac{7+\cos 4 x}{8} \Rightarrow \frac{3}{4} \leq A \leq 1$
3. Q3JEE Main 2026 (05 Apr, Shift 2)Solving Trigonometric Equation
   If $S = \left\{\theta \in [-\pi, \pi] : \cos\theta \cos\dfrac{5\theta}{2} = \cos 7\theta \cos\dfrac{7\theta}{2}\right\}$, then $n(S)$ is equal to _______.
   Show answer & solution

   Answer: 19

   Given equation: $\cos\theta \cos\dfrac{5\theta}{2} = \cos 7\theta \cos\dfrac{7\theta}{2}$ Multiplying both sides by $2$: $2\cos\theta \cos\dfrac{5\theta}{2} = 2\cos 7\theta \cos\dfrac{7\theta}{2}$ Using the identity $2\cos A \cos B = \cos(A+B) + \cos(A-B)$: $\cos\left(\theta + \dfrac{5\theta}{2}\right) + \cos\left(\theta - \dfrac{5\theta}{2}\right) = \cos\left(7\theta + \dfrac{7\theta}{2}\right) + \cos\left(7\theta - \dfrac{7\theta}{2}\right)$ $\cos\left(\dfrac{7\theta}{2}\right) + \cos\left(\dfrac{-3\theta}{2}\right) = \cos\left(\dfrac{21\theta}{2}\right) + \cos\left(\dfrac{7\theta}{2}\right)$ Since $\cos(-x) = \cos x$, we get: $\cos\left(\dfrac{7\theta}{2}\right) + \cos\left(\dfrac{3\theta}{2}\right) = \cos\left(\dfrac{21\theta}{2}\right) + \cos\left(\dfrac{7\theta}{2}\right)$ $\cos\left(\dfrac{3\theta}{2}\right) = \cos\left(\dfrac{21\theta}{2}\right)$ $\cos\left(\dfrac{21\theta}{2}\right) - \cos\left(\dfrac{3\theta}{2}\right) = 0$ Using the identity $\cos C - \cos D = -2\sin\left(\dfrac{C+D}{2}\right)\sin\left(\dfrac{C-D}{2}\right)$: $-2\sin\left(\dfrac{\dfrac{21\theta}{2} + \dfrac{3\theta}{2}}{2}\right)\sin\left(\dfrac{\dfrac{21\theta}{2} - \dfrac{3\theta}{2}}{2}\right) = 0$ $-2\sin(6\theta)\sin\left(\dfrac{9\theta}{2}\right) = 0$ This gives $\sin(6\theta) = 0$ or $\sin\left(\dfrac{9\theta}{2}\right) = 0$. Case 1: $\sin(6\theta) = 0$ $6\theta = n\pi \Rightarrow \theta = \dfrac{n\pi}{6}$ for $n \in \mathbb{Z}$. Since $\theta \in [-\pi, \pi]$, we have $-\pi \le \dfrac{n\pi}{6} \le \pi \Rightarrow -6 \le n \le 6$. The possible values for $n$ are $\{-6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6\}$. This gives $13$ solutions. Case 2: $\sin\left(\dfrac{9\theta}{2}\right) = 0$ $\dfrac{9\theta}{2} = m\pi \Rightarrow \theta = \dfrac{2m\pi}{9}$ for $m \in \mathbb{Z}$. Since $\theta \in [-\pi, \pi]$, we have $-\pi \le \dfrac{2m\pi}{9} \le \pi \Rightarrow -4.5 \le m \le 4.5$. The possible values for $m$ are $\{-4, -3, -2, -1, 0, 1, 2, 3, 4\}$. This gives $9$ solutions. To find common solutions, we equate the two sets of solutions: $\dfrac{n\pi}{6} = \dfrac{2m\pi}{9} \Rightarrow 3n = 4m$ For $m \in \{-4, -3, -2, -1, 0, 1, 2, 3, 4\}$, $m$ must be a multiple of $3$. The valid values for $m$ are $-3, 0, 3$, which correspond to $n = -4, 0, 4$. Thus, there are $3$ common solutions. Total number of unique solutions $n(S) = 13 + 9 - 3 = 19$. Answer: $19$
4. Q4JEE Main 2026 (02 Apr, Shift 1)Solving Trigonometric Equation
   Let $S = \{x \in [-\pi, \pi] : \sin x (\sin x + \cos x) = a, a \in \mathbb{Z}\}$. Then $n(S)$ is equal to :
   1. A.$3$
   2. B.$6$
   3. C.$7$
   4. D.$9$
   Show answer & solution

   Answer: (D)

   Given the equation: $\sin x (\sin x + \cos x) = a$ $\Rightarrow \sin^2 x + \sin x \cos x = a$ Using the double angle formulas $\sin^2 x = \dfrac{1 - \cos 2x}{2}$ and $\sin x \cos x = \dfrac{\sin 2x}{2}$, we get: $\dfrac{1 - \cos 2x}{2} + \dfrac{\sin 2x}{2} = a$ $\Rightarrow \sin 2x - \cos 2x = 2a - 1$ We know that the range of $A\sin \theta + B\cos \theta$ is $[-\sqrt{A^2+B^2}, \sqrt{A^2+B^2}]$. Thus, the range of $\sin 2x - \cos 2x$ is $[-\sqrt{2}, \sqrt{2}]$. $\Rightarrow -\sqrt{2} \le 2a - 1 \le \sqrt{2}$ $\Rightarrow 1 - 1.414 \le 2a \le 1 + 1.414$ $\Rightarrow -0.414 \le 2a \le 2.414$ $\Rightarrow -0.207 \le a \le 1.207$ Since $a \in \mathbb{Z}$, the possible values for $a$ are $0$ and $1$. Case 1: $a = 0$ $\sin x (\sin x + \cos x) = 0$ $\Rightarrow \sin x = 0$ or $\sin x + \cos x = 0$ For $x \in [-\pi, \pi]$, $\sin x = 0 \Rightarrow x \in \{-\pi, 0, \pi\}$ (3 solutions) $\sin x + \cos x = 0 \Rightarrow \tan x = -1 \Rightarrow x \in \{-\dfrac{\pi}{4}, \dfrac{3\pi}{4}\}$ (2 solutions) Total solutions for $a = 0$ is $5$. Case 2: $a = 1$ $\sin^2 x + \sin x \cos x = 1$ $\Rightarrow \sin x \cos x = 1 - \sin^2 x$ $\Rightarrow \sin x \cos x = \cos^2 x$ $\Rightarrow \cos x (\sin x - \cos x) = 0$ $\Rightarrow \cos x = 0$ or $\sin x - \cos x = 0$ For $x \in [-\pi, \pi]$, $\cos x = 0 \Rightarrow x \in \{-\dfrac{\pi}{2}, \dfrac{\pi}{2}\}$ (2 solutions) $\sin x - \cos x = 0 \Rightarrow \tan x = 1 \Rightarrow x \in \{-\dfrac{3\pi}{4}, \dfrac{\pi}{4}\}$ (2 solutions) Total solutions for $a = 1$ is $4$. The total number of elements in the set $S$ is $n(S) = 5 + 4 = 9$. Answer: $9$
5. Q5JEE Main 2026 (24 Jan, Shift 2)Solving Trigonometric Equation
   The number of elements in the set $\left\{x \in\left[0,180^{\circ}\right]: \tan \left(x+100^{\circ}\right)=\tan \left(x+50^{\circ}\right) \tan x \tan \left(x-50^{\circ}\right)\right\}$ is $\_\_\_\_$.
   Show answer & solution

   Answer: 4

   $\frac{\tan(x+100°)}{\tan x} = \tan(x+50°)\tan(x-50°)$ $\frac{\sin(x+100°)\cos x}{\cos(x+100°)\sin x} = \frac{\sin(x+50°)\sin(x-50°)}{\cos(x+50°)\cos(x-50°)}$ Applying Componendo & Dividendo: $\frac{\sin(2x+100°)}{\sin 100°} = \frac{\cos 100°}{-\cos 2x}$ $2\sin(2x+100°)\cos 2x + \sin 200° = 0$ $\sin(4x+100°) + \sin 100° + \sin 200° = 0$ $\sin(4x+100°) = -2\sin 150° \cos 50°$ $\sin(4x+100°) = -\cos 50° = \sin(-40°)$ $\therefore 4x + 100° = n\pi + (-1)^n(-40°)$ $x = \frac{n\pi + (-1)^{n+1}(40°) - 100°}{4}$ $\therefore x = 30°, 55°, 120°, 145°$ in $(0, \pi)$ $\therefore$ Number of solutions $= 4$