Trigonometric Equations JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Trigonometric Equations, free to read — no sign-in needed. The full chapter has 25 questions; sign in to attempt the remaining 20 in the exam simulator.

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1. Q1JEE Advanced Adv 2026 (Paper 2)
   Passage: Consider the curve $C_1$ given by $y = e^{-x}$ for $x \in [0, 10\pi]$, and the curve $C_2$ given by $y = e^{-x}(\sin x + \cos x)$ for $x \in [0, 10\pi]$. Let $n$ be the total number of points of intersection of the curves $C_1$ and $C_2$. Suppose that $\alpha_1, \alpha_2, \ldots, \alpha_n \in [0, 10\pi]$ are the $x$-coordinates of the points of intersection of the curves $C_1$ and $C_2$ such that $\alpha_1 \lt \alpha_2 \lt \cdots \lt \alpha_n$. Question: The value of $n$ is ___________.
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   Answer: 11

   Equating the equations of the curves $C_1$ and $C_2$: $e^{-x} = e^{-x}(\sin x + \cos x)$ Since $e^{-x} \neq 0$ for all real $x$, dividing both sides by $e^{-x}$ gives: $\sin x + \cos x = 1$ Multiplying both sides by $\dfrac{1}{\sqrt{2}}$: $\dfrac{1}{\sqrt{2}}\sin x + \dfrac{1}{\sqrt{2}}\cos x = \dfrac{1}{\sqrt{2}}$ $\sin\left(x + \dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt{2}}$ The general solution is given by: $x + \dfrac{\pi}{4} = 2m\pi + \dfrac{\pi}{4}$ or $x + \dfrac{\pi}{4} = 2m\pi + \dfrac{3\pi}{4}$ for integer $m$ This simplifies to: $x = 2m\pi$ or $x = 2m\pi + \dfrac{\pi}{2}$ Given the interval $x \in [0, 10\pi]$, the valid values for $x$ are: For $x = 2m\pi$: $x = 0, 2\pi, 4\pi, 6\pi, 8\pi, 10\pi$ (6 solutions) For $x = 2m\pi + \dfrac{\pi}{2}$: $x = \dfrac{\pi}{2}, \dfrac{5\pi}{2}, \dfrac{9\pi}{2}, \dfrac{13\pi}{2}, \dfrac{17\pi}{2}$ (5 solutions) The total number of points of intersection is $n = 6 + 5 = 11$. Answer: $11$
2. Q2JEE Advanced Adv 2026 (Paper 1)
   Match each entry in List-I to the correct entry in List-II and choose the correct option. List-IList-II(P) The number of elements in the set $\{x \in [-\pi, \pi] : \sin^6 x + \cos^4 x = 1\}$(1) is $1$(Q) The number of elements in the set $\left\{x \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right] : \sin^2 x + \cos^6 x = 1\right\}$(2) is $2$(R) The number of elements in the set $\left\{x \in [-\pi, \pi] : \cos^2\left(\dfrac{x}{2}\right) - \sin^2 x = \dfrac{1}{2}\right\}$(3) is $3$(S) The number of elements in the set $\left\{x \in [-2\pi, 2\pi] : 6\sin^2\left(\dfrac{x}{2}\right) - \cos 3x = 3\right\}$(4) is $4$(5) is $5$
   1. A.(P) $\rightarrow$ (2), (Q) $\rightarrow$ (5), (R) $\rightarrow$ (3), (S) $\rightarrow$ (4)
   2. B.(P) $\rightarrow$ (5), (Q) $\rightarrow$ (3), (R) $\rightarrow$ (2), (S) $\rightarrow$ (4)
   3. C.(P) $\rightarrow$ (5), (Q) $\rightarrow$ (4), (R) $\rightarrow$ (1), (S) $\rightarrow$ (3)
   4. D.(P) $\rightarrow$ (4), (Q) $\rightarrow$ (3), (R) $\rightarrow$ (2), (S) $\rightarrow$ (5)
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   Answer: (B)

   For (P): $\sin^6 x + \cos^4 x = 1$ $\sin^6 x + (1 - \sin^2 x)^2 = 1$ $\sin^6 x + 1 - 2\sin^2 x + \sin^4 x = 1$ $\sin^2 x (\sin^4 x + \sin^2 x - 2) = 0$ $\sin^2 x (\sin^2 x - 1)(\sin^2 x + 2) = 0$ This gives $\sin^2 x = 0$ or $\sin^2 x = 1$. In the interval $[-\pi, \pi]$, the solutions are $x \in \{-\pi, -\dfrac{\pi}{2}, 0, \dfrac{\pi}{2}, \pi\}$. Number of solutions = $5$. For (Q): $\sin^2 x + \cos^6 x = 1$ $1 - \cos^2 x + \cos^6 x = 1$ $\cos^2 x (\cos^4 x - 1) = 0$ This gives $\cos^2 x = 0$ or $\cos^2 x = 1$. In the interval $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$, the solutions are $x \in \left\{-\dfrac{\pi}{2}, 0, \dfrac{\pi}{2}\right\}$. Number of solutions = $3$. For (R): $\cos^2\left(\dfrac{x}{2}\right) - \sin^2 x = \dfrac{1}{2}$ $\dfrac{1 + \cos x}{2} - (1 - \cos^2 x) = \dfrac{1}{2}$ $1 + \cos x - 2 + 2\cos^2 x = 1$ $2\cos^2 x + \cos x - 2 = 0$ Solving the quadratic equation for $\cos x$, we get: $\cos x = \dfrac{-1 \pm \sqrt{17}}{4}$ Since $\cos x \in [-1, 1]$, we reject the negative root. Thus, $\cos x = \dfrac{\sqrt{17} - 1}{4}$. Since $0 \lt \dfrac{\sqrt{17} - 1}{4} \lt 1$, there are exactly $2$ solutions in $[-\pi, \pi]$. Number of solutions = $2$. For (S): $6\sin^2\left(\dfrac{x}{2}\right) - \cos 3x = 3$ $3(1 - \cos x) - (4\cos^3 x - 3\cos x) = 3$ $3 - 3\cos x - 4\cos^3 x + 3\cos x = 3$ $-4\cos^3 x = 0 \Rightarrow \cos x = 0$ In the interval $[-2\pi, 2\pi]$, the solutions are $x \in \left\{-\dfrac{3\pi}{2}, -\dfrac{\pi}{2}, \dfrac{\pi}{2}, \dfrac{3\pi}{2}\right\}$. Number of solutions = $4$. Therefore, the correct matching is (P) $\rightarrow$ (5), (Q) $\rightarrow$ (3), (R) $\rightarrow$ (2), (S) $\rightarrow$ (4). Answer: (P) $\rightarrow$ (5), (Q) $\rightarrow$ (3), (R) $\rightarrow$ (2), (S) $\rightarrow$ (4)
3. Q3JEE Advanced Adv 2024 (Paper 2)
   Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function defined by $f(x)=\left\{\begin{array}{cc}x^2 \sin \left(\frac{\pi}{x^2}\right), & \text { if } x \neq 0 \\ 0, & \text { if } x=0\end{array}\right.$ Then which of the following statements is TRUE?
   1. A.$f(x)=0$ has infinitely many solutions in the interval $\left[\frac{1}{10^{10}}, \infty\right)$.
   2. B.$f(x)=0$ has no solutions in the interval $\left[\frac{1}{\pi}, \infty\right)$.
   3. C.The set of solutions of $f(x)=0$ in the interval $\left(0, \frac{1}{10^{10}}\right)$ is finite
   4. D.$f(x)=0$ has more than 25 solutions in the interval $\left(\frac{1}{\pi^2}, \frac{1}{\pi}\right)$.
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   Answer: (D)

   Option-1 : $\mathrm{f}(\mathrm{x})=\mathrm{x}^2 \sin \frac{\pi}{\mathrm{x}^2}$ $f(x)=0 \Rightarrow \sin \frac{\pi}{x^2}=0 \Rightarrow \frac{\pi}{x^2}=n \pi, n \in N$ $\mathrm{x}^2=\frac{1}{\mathrm{n}} \Rightarrow \mathrm{x}=\frac{1}{\sqrt{\mathrm{n}}}$ $\frac{1}{\sqrt{\mathrm{n}}} \geq \frac{1}{10^{10}} \Rightarrow 10^{10} \geq \sqrt{\mathrm{n}} \Rightarrow \mathrm{n} \leq 10^{20}$, finite number of solutions Option-2 : $\mathrm{x}=\frac{1}{\sqrt{\mathrm{n}}} \Rightarrow \frac{1}{\sqrt{\mathrm{n}}}\gt \frac{1}{\pi} \Rightarrow \pi\gt \sqrt{\mathrm{n}} \Rightarrow \mathrm{n} \lt \pi^2$, Number of solutions is 9 Option-3 $: \mathrm{x}=\frac{1}{\sqrt{\mathrm{n}}}, \frac{1}{\sqrt{\mathrm{n}}} \lt \frac{1}{10^{10}} \Rightarrow \sqrt{\mathrm{n}}\gt 10^{10} \Rightarrow \mathrm{n}\gt 10^{20}$, Infinite number of solutions Option-4 : $\frac{1}{\pi^2} \lt \frac{1}{\sqrt{\mathrm{n}}} \lt \frac{1}{\pi} \Rightarrow \sqrt{\mathrm{n}} \in\left(\pi, \pi^2\right) \Rightarrow \mathrm{n} \in\left(\pi^2, \pi^4\right)$, Definitely more than 25 solutions
4. Q4JEE Advanced Adv 2024 (Paper 1)
   Let $\frac{\pi}{2} \lt x \lt \pi$ be such that $\cot x=\frac{-5}{\sqrt{11}}$. Then $\left(\sin \frac{11 x}{2}\right)(\sin 6 x-\cos 6 x)+\left(\cos \frac{11 x}{2}\right)(\sin 6 x+\cos 6 x)$ is equal to
   1. A.$\frac{\sqrt{11}-1}{2 \sqrt{3}}$
   2. B.$\frac{\sqrt{11}+1}{2 \sqrt{3}}$
   3. C.$\frac{\sqrt{11}+1}{3 \sqrt{2}}$
   4. D.$\frac{\sqrt{11}-1}{3 \sqrt{2}}$
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   Answer: (B)

   Given, $\begin{aligned} & \cot x=-\frac{5}{\sqrt{11}} \\ & \frac{1-\tan ^2 \frac{x}{2}}{2 \tan \frac{x}{2}}=-\frac{5}{\sqrt{11}} \\ & \tan \frac{x}{2}=\sqrt{11},-\frac{1}{\sqrt{11}} \\ & \text{So, } \tan \frac{x}{2}=\sqrt{11}, \text { As } \frac{\pi}{4} \lt \frac{x}{2} \lt \frac{\pi}{2}\end{aligned}$ $\begin{aligned} & \text{As, } x \in\left(\frac{\pi}{2}, \pi\right) \\ & \text{Now, } \left(\sin \frac{11 x}{2}\right)(\sin 6 x-\cos 6 x)+\left(\cos \frac{11 x}{2}\right)(\sin 6 x+\cos 6 x) \\ & =\left\{\sin 6 x \sin \frac{11 x}{2}+\cos \frac{11 x}{2} \cos 6 x \right\} \\ & =\cos \left(6 x-\frac{11 x}{2}\right)+\sin \left(6 x-\frac{11 x}{2}\right) \\ & =\cos \frac{x}{2}+\sin \frac{x}{2} \\ & =\frac{1}{2 \sqrt{3}}+\frac{\sqrt{11}}{2 \sqrt{3}} \\ & =\frac{\sqrt{11}+1}{2 \sqrt{3}} \Rightarrow \text { Option (2) is correct. }\end{aligned}$
5. Q5JEE Advanced Adv 2023 (Paper 2)
   For any $y\in ℝ$, let ${\cot }^{-1}\left(y\right)\in \left(0,\pi \right)$ and ${\tan }^{-1}\left(y\right)\in \left(-\dfrac{\pi }{2},\dfrac{\pi }{2}\right)$. Then the sum of all the solutions of the equation ${\tan }^{-1}\left(\dfrac{6y}{9-{y}^{2}}\right)+{\cot }^{-1}\left(\dfrac{9-{y}^{2}}{6y}\right)=\dfrac{2\pi }{3}$ for $0\lt |y|\lt 3$, is equal to
   1. A.$2\sqrt{3}-3$
   2. B.$3-2\sqrt{3}$
   3. C.$4\sqrt{3}-6$
   4. D.$6-4\sqrt{3}$
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   Answer: (C)

   Given, ${\tan }^{-1}\left(\dfrac{6y}{9-{y}^{2}}\right)+{\cot }^{-1}\left(\dfrac{9-{y}^{2}}{6y}\right)=\dfrac{2\pi }{3}$ And $0\lt |y|\lt 3\Rightarrow y\in \left(-3,3\right)-\left\{0\right\}$ Now taking, Case-l: When $\dfrac{6y}{9-{y}^{2}}\gt 0\Rightarrow y\gt 0$ ${\tan }^{-1}⁡\left(\dfrac{6y}{9-{y}^{2}}\right)+{\tan }^{-1}⁡\left(\dfrac{6y}{9-{y}^{2}}\right)=\dfrac{2\pi }{3}$ $\Rightarrow 2{\tan }^{-1}\left(\dfrac{6y}{9-{y}^{2}}\right)=\dfrac{2\pi }{3}$ $\Rightarrow {\tan }^{-1}\left(\dfrac{6y}{9-{y}^{2}}\right)=\dfrac{\pi }{3}$ $\Rightarrow \dfrac{6y}{9-{y}^{2}}=\sqrt{3}$ $\Rightarrow 6y=9\sqrt{3}-\sqrt{3}{y}^{2}$ $\Rightarrow \sqrt{3}{y}^{2}+6y-9\sqrt{3}=0$ $\Rightarrow \sqrt{3}{y}^{2}+9y-3y-9\sqrt{3}=0$ $\Rightarrow \left(y+3\sqrt{3}\right)\left(\sqrt{3}y-3\right)=0$ $y\neq -3\sqrt{3}∴y=\sqrt{3}$ as $y\in (0,3)$ Now taking, Case-II: When $\dfrac{6y}{9-{y}^{2}}\lt 0\Rightarrow y\lt 0$ ${\tan }^{-1}\left(\dfrac{6y}{9-{y}^{2}}\right)+\pi +{\tan }^{-1}\left(\dfrac{6y}{9-{y}^{2}}\right)=\dfrac{2\pi }{3}$ $\Rightarrow 2{\tan }^{-1}\left(\dfrac{6y}{9-{y}^{2}}\right)=-\dfrac{\pi }{3}$ $\Rightarrow {\tan }^{-1}\left(\dfrac{6y}{9-{y}^{2}}\right)=-\dfrac{\pi }{6}$ $\Rightarrow \dfrac{6y}{9-{y}^{2}}=-\dfrac{1}{\sqrt{3}}$ $\Rightarrow 6\sqrt{3}y=-9+{y}^{2}$ $\Rightarrow {y}^{2}-6\sqrt{3}y-9=0$ $\Rightarrow y=\dfrac{6\sqrt{3}\pm \sqrt{108+36}}{2}=\dfrac{6\sqrt{3}\pm 12}{2}=3\sqrt{3}\pm 6$ $\text{As}y\in (-3.0),\text{so}y=3\sqrt{3}-6$ $∴\text{Sum of solutions}=\sqrt{3}+(3\sqrt{3}-6)=4\sqrt{3}-6$