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JEE • MATHEMATICS • TRIGONOMETRIC EQUATIONS

Trigonometric equations practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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31 Easy
31 of 581
Application +4 | -1
If tan⁡mθ+cot⁡nθ=0\tan m \theta+\cot n \theta=0, then the general value of θ\theta is
View answer and explanation
Correct option: A
(2r+1)π2(m−n)\frac{(2 r+1) \pi}{2(m-n)}
Option B ((2r+1)π2(m+n)\frac{{(2 r+1) \pi}}{{2(m+n)}}) is correct because by rewriting the equation in terms of tangent and using the general solution for tangent, we can find the general value of theta.
32 Easy
32 of 581
Analyzing +4 | -1
The largest positive solution of 1+sin⁡4x=cos⁡23x1 + {\sin ^4}x = {\cos ^2}3x in [− 5π/2,   5π/2]\left[ { - \,5\pi /2,\;\,5\pi /2} \right] is

View answer and explanation
Correct option: B
2  π2\;\pi
Option B (2  π2\;\pi) is correct because it is the largest positive solution within the given interval that satisfies both sin(3x) = 0 and sin(x) = 0.
33 Easy
33 of 581
Application +4 | -1
If tan⁡  (cot⁡  x)=cot⁡  (tan⁡  x)\tan \,\,\left( {\cot \,\,x} \right) = \cot \,\,\left( {\tan \,\,x} \right), then sin⁡  2x\sin \,\,2x is equal to
View answer and explanation
Correct option: B
4(2n+1)π\frac{4}{{\left( {2n + 1} \right)\pi }}
Option B (4(2n+1)π\frac{4}{{\left( {2n + 1} \right)\pi }}) is correct because by converting cotangent to tangent and using trigonometric identities, we can solve for sin⁡2x\sin 2x in terms of nn.
34 Easy
34 of 581
Analyzing +4 | -1
Number of solutions of the equation sin⁡  2θ+2=4  sin⁡  θ+cos⁡  θ\sin \,\,2\theta + 2 = 4\,\,\sin \,\,\theta + \cos \,\,\theta lying in the interval [π,  5  π]\left[ {\pi ,\,\,5\;\pi } \right], is
View answer and explanation
Correct option: C
4\mathrm{4}
Option C (4) is correct because solving the trigonometric equation yields four solutions within the specified interval.
35 Easy
35 of 581
Understanding +4 | -1
Consider the following statements.

I. Inequalities involving the symbol ≥\geq or ≤\leq are called slack inequalities.

II. Inequalities which do not involve variables are called numerical inequalities.

Choose the correct option.
View answer and explanation
Correct option: C
Both are true
Option C (Both are true) is correct because both statements accurately describe slack and numerical inequalities.
36 Medium
36 of 581
Application +4 | -1
If x=sin⁡2θcos⁡θx=\sin ^2 \theta \cos \theta and y=cos⁡2θsin⁡θy=\cos ^2 \theta \sin \theta, then which of the following is true ? 0<θ<π2,x>y>00 < \theta < \frac{\pi}{2}, x > y > 0
View answer and explanation
Correct option: B
(x2y)23+(y2x)23=1\left(\frac{x^2}{y}\right)^{\frac{2}{3}}+\left(\frac{y^2}{x}\right)^{\frac{2}{3}}=1
Option B ((x2y)23+(y2x)23=1\left(\frac{{x^2}}{{y}}\right)^{\frac{{2}}{{3}}}+\left(\frac{{y^2}}{{x}}\right)^{\frac{{2}}{{3}}}=1) is correct because it is the result of simplifying the given equations using trigonometric identities.
37 Medium
37 of 581
Application +4 | -1
The value of 3[sin⁡4(3π2−α)+sin⁡4(3π+α)]3\left[\sin ^4\left(\frac{3 \pi}{2}-\alpha\right)+\sin ^4(3 \pi+\alpha)\right] −2[sin⁡6(π2+α)+sin⁡6(5π−α)]-2\left[\sin ^6\left(\frac{\pi}{2}+\alpha\right)+\sin ^6(5 \pi-\alpha)\right] is
View answer and explanation
Correct option: B
1
The correct answer is 1 because the expression simplifies to 1 using trigonometric identities and reduction formulas.
38 Easy
38 of 581
Application +4 | -1
The general value of θ\mathrm{\theta } satisfying sin⁡2  θ+sin⁡  θ=2{\sin ^2}\,\,\mathrm{\theta } + \sin \,\,\mathrm{\theta } = 2 is
View answer and explanation
Correct option: C
nπ  +  (−1)n  π2n\pi \,\, + \,\,{\left( { - 1} \right)^n}\,\,\frac{\pi }{2}
Option C (nπ,+,(−,1)nπ2n\pi \\, + \\,{{\left( { - \\,1} \right)}^n}\frac{\pi }{2}) is correct because it represents the general solution to the equation sin⁡2θ+sin⁡θ=2\sin^2 \theta + \sin \theta = 2.
39 Medium
39 of 581
Application +4 | -1
The solutions of the equation 4  cos⁡2  x+6  sin⁡2  x=54\,\,{\cos ^2}\,\,x + 6\,\,{\sin ^2}\,\,x = 5 are
View answer and explanation
Correct option: A
x=nπ  ±  π4x = n\pi \,\, \pm \,\,\frac{\pi }{4}
Option A (x=nπ ± π4x = n\pi \, \pm \, \frac{\pi }{4}) is correct because solving the trigonometric equation 4cos⁡2x+6sin⁡2x=54{\cos ^2}x + 6{\sin ^2}x = 5 leads to sin⁡x=±12\sin x = \pm \frac{1}{\sqrt{2}}, which gives the general solution x=nπ±π4x = n\pi \pm \frac{\pi}{4}.
40 Easy
40 of 581
Application +4 | -1
Number of integral value(s) of mm for which the equation sin⁡x−3cos⁡x=4m−64−m\sin x-\sqrt{3} \cos x=\frac{4 m-6}{4-m} has solutions, x∈[0,2π]x \in[0,2 \pi], is___.
View answer and explanation
Correct option: A
4
Option A (4) is correct because the number of integral values of m for which the equation has solutions is 4, derived from the inequality condition.
41 Medium
41 of 581
Application +4 | -1
sin⁡3θ2cos⁡2θ+1=12\frac{\sin 3 \theta}{2 \cos 2 \theta+1}=\frac{1}{2} if -
View answer and explanation
Correct option: C
θ=nπ+(−1)nπ6\theta={n} \pi+(-1)^{{n}} \frac{\pi}{6}
Option C (θ=nπ+(−1)nπ6\theta={n} \pi+(-1)^{{n}} \frac{\pi}{6}) is correct because it represents the general solution of the simplified trigonometric equation sin⁡θ=12\sin \theta = \frac{1}{2}.
42 Medium
42 of 581
Analyzing +4 | -1
If the expression nsin⁡2θ+2ncos⁡(θ+α)sin⁡αsin⁡θn \sin ^2 \theta+2 n \cos (\theta+\alpha) \sin \alpha \sin \theta +cos⁡2(α+θ)+\cos 2(\alpha+\theta) is independent of θ\theta, then the value of nn is
View answer and explanation
Correct option: B
2
Option A (2) is correct because setting n=2 makes the expression independent of theta, as the theta-dependent terms cancel out.
43 Medium
43 of 581
Application +4 | -1
The solution set of (2cos⁡x−1)  (3+2 cos⁡x)=0\left( {2\cos x - 1} \right)\,\,\left( {3 + 2\,\cos x} \right) = 0 in the interval 0≤x≤2  π0 \le x \le 2\;\pi, is
View answer and explanation
Correct option: B
{π3,  5  π3}\left\{ {\frac{\pi }{3},\,\,\frac{{5\;\pi }}{3}} \right\}
Option B ({π3,  5  π3}\left\{ {\frac{\pi }{3},\,\,\frac{{5\;\pi }}{3}} \right\}) is correct because these are the only solutions to the equation within the given interval.
44 Easy
44 of 581
Analyzing +4 | -1
The general solution of the equation cos⁡  x   cos   6x=−1\cos \,\,x\,\;\mathrm{cos}\,\;6x = - 1 is
View answer and explanation
Correct option: A
x=(2n+1)π,  n∈Zx = \left( {2n + 1} \right)\pi ,\,\,n \in Z
Option A is correct because the general solution is derived by setting both cos(x) and cos(6x) to -1 and finding the common solution.
45 Medium
45 of 581
Application +4 | -1
If sin⁡x+3cos⁡x=2\sin x+\sqrt{3} \cos x=\sqrt{2}, then xx is equal to
View answer and explanation
Correct option: B
2nπ+5π12,2nπ−π122 n \pi+\frac{5 \pi}{12}, 2 n \pi-\frac{\pi}{12}
Option B (2nπ+5π12,2nπ−π122 n \pi+\frac{{5 \pi}}{{12}}, 2 n \pi-\frac{{\pi}}{{12}}) is correct because it represents the general solution for x after solving the trigonometric equation.
46 Easy
46 of 581
Analyzing +4 | -1
The number of solutions of the equation sin⁡x=cos⁡  3x    in   [0,  π]\sin x = \cos \,\,3x\,\,\;in\;\,\left[ {0,\,\,\pi } \right], is
View answer and explanation
Correct option: C
3\mathrm{3}
Option C (3) is correct because the equation sin⁡x=cos⁡3x\sin x = \cos 3x has three solutions in the interval [0,π][0, \pi].
47 Easy
47 of 581
Application +4 | -1
The equation sin⁡xcos⁡x=2\sin x \cos x=2 has
View answer and explanation
Correct option: D
No solutions
Option D (No solutions) is correct because the range of sin⁡xcos⁡x\sin x \cos x is [−1/2,1/2][-1/2, 1/2], and the equation sin⁡xcos⁡x=2\sin x \cos x = 2 has no solution since 2 is outside this range.
48 Medium
48 of 581
Application +4 | -1
If sin⁡xsin⁡y=12,cos⁡xcos⁡y=32\frac{\sin x}{\sin y}=\frac{1}{2}, \frac{\cos x}{\cos y}=\frac{3}{2}, where x,y∈(0,π2)x, y \in\left(0, \frac{\pi}{2}\right), then the value of tan⁡(x+y)\tan (x+y) is
View answer and explanation
Correct option: D
15\sqrt{15}
Option D (15\sqrt{15}) is correct because by using the given ratios and trigonometric identities, we can find the value of tan⁡x\tan x and then calculate tan⁡(x+y)\tan(x+y).
49 Medium
49 of 581
Understanding +4 | -1
The equation asin⁡x+bcos⁡x=ca \sin x+b \cos x=c, where ∣c∣>a2+b2|c|>\sqrt{a^2+b^2} has
View answer and explanation
Correct option: C
no solution
Option B (0) is correct because lim⁡x→0xsin⁡1x=0\lim_{x \rightarrow 0} x \sin \frac{1}{x} = 0 due to the Squeeze Theorem.
50 Easy
50 of 581
Application +4 | -1
Let Δ=∣1sin⁡θ1−sin⁡θ1sin⁡θ−1−sin⁡θ1∣\Delta=\left|\begin{array}{ccc}1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1\end{array}\right|, then Δ\Delta lies in the interval.
View answer and explanation
Correct option: C
[1,4][1,4]
Option C ([1,4]) is correct because by expanding the determinant and simplifying, the value of delta lies in the interval [2,4], which is a subset of [1,4].
51 Medium
51 of 581
Analyzing +4 | -1
The most general solution of

21+ ∣ cos⁡x ∣+ cos⁡2x+ ∣ cos⁡3x ∣+…∞=4{2^{1 + \,\left| {\,\cos x\,} \right| + \,{{\cos }^2}x + \,\left| {\,{{\cos }^3}x\,} \right| + \ldots \infty }} = 4 is given by
View answer and explanation
Correct option: A
x=nπ  ±  π3,  n∈Zx = n\pi \,\, \pm \,\,\frac{\pi }{3},\,\,n \in Z
Option A (x = nπ ± π/3, n ∈ Z) is correct because solving the equation leads to cos x = ±1/2, which gives the general solution x = nπ ± π/3.
52 Medium
52 of 581
Analyzing +4 | -1
Let f(θ)=3(sin⁡4(3π2−θ)+sin⁡4(3π+θ))−2(1−sin⁡22θ)f(\theta)=3\left(\sin ^{4}\left(\frac{3 \pi}{2}-\theta\right)+\sin ^{4}(3 \pi+\theta)\right)-2\left(1-\sin ^{2} 2 \theta\right) and S={θ∈[0,π]:f′(θ)=−32}S=\left\{\theta \in[0, \pi]: f^{\prime}(\theta)=-\frac{\sqrt{3}}{2}\right\}. If 4β=∑θ∈Sθ4 \beta=\sum_{\theta \in S} \theta, then f(β)f(\beta) is equal to
View answer and explanation
Correct option: C
54\frac{5}{4}
The correct answer is C (54\frac{5}{4}). This involves simplifying the trigonometric function, finding its derivative, solving the equation f′(θ)=−32f'(\theta) = -\frac{\sqrt{3}}{2}, and then evaluating f(β)f(\beta).
53 Medium
53 of 581
Analyzing +4 | -1
If r>0,−π≤θ≤πr > 0,-\pi \leq \theta \leq \pi and r,θr, \theta satisfy rsin⁡θ=3r \sin \theta=3 and r=4(1+sin⁡θ)r=4(1+\sin \theta), then the number of possible solutions of the pair (r,θ)(r, \theta) is
View answer and explanation
Correct option: A
2
Option A (2) is correct because there are two possible solutions for the pair (r,θ)(r, \theta) after solving the system of equations.
54 Medium
54 of 581
Analyzing +4 | -1
Statement-1: The number of common solutions of the trigonometric equations 2sin⁡2θ−cos⁡2θ=02 \sin ^{2} \theta-\cos 2 \theta=0 and 2cos⁡2θ−3sin⁡θ=02 \cos ^{2} \theta-3 \sin \theta=0 in the interval [0,2π][0,2 \pi] is two.

Statement-2: The number of solutions of the equation, 2 cos⁡2θ−3sin⁡θ=0\cos ^{2} \theta-3 \sin \theta=0 in the interval [0,π][0, \pi] is two.
View answer and explanation
Correct option: B
Statement-1 is true; Statement-2 is true; Statement-2 is not a correct explanation for statement-1.
Statement-1 is true because the equations have two common solutions, and Statement-2 is true because the equation has two solutions in the interval, but Statement-2 doesn't explain Statement-1.
55 Easy
55 of 581
Application +4 | -1
If sin⁡θ+cosec⁡θ=2\sin \theta+\operatorname{cosec} \theta=2, then sin⁡nθ+cosec⁡nθ\sin ^n \theta+\operatorname{cosec}^n \theta is equal to
View answer and explanation
Correct option: A
2
Option A (2) is correct because if sin(theta) + cosec(theta) = 2, then sin(theta) = 1, and therefore sin^n(theta) + cosec^n(theta) = 1 + 1 = 2.
56 Medium
56 of 581
Application +4 | -1
If 0<β<α<π4,cos⁡(α+β)=350 < \beta < \alpha < \frac{\pi}{4}, \cos (\alpha+\beta)=\frac{3}{5} and cos⁡(α−β)=45\cos (\alpha-\beta)=\frac{4}{5}, then sin⁡2α\sin 2 \alpha is equal to
View answer and explanation
Correct option: B
1
Option B (1) is correct because using the formula sin⁡2α=sin⁡(((α+β)+(α−β)))\sin 2\alpha = \sin(((\alpha + \beta) + (\alpha - \beta))) and the given values, we get sin⁡2α=1\sin 2\alpha = 1.
57 Easy
57 of 581
Application +4 | -1
If 1+sin⁡θ+sin⁡2θ+……∞=4+23,0<θ<π,θ≠π21+\sin \theta+\sin ^2 \theta+\ldots \ldots \infty=4+2 \sqrt{3}, 0 < \theta < \pi, \theta \neq \frac{\pi}{2}, then
View answer and explanation
Correct option: D
θ=π3\theta=\frac{\pi}{3} or 2π3\frac{2 \pi}{3}
Option D (θ=π3\theta=\frac{\pi}{3} or 2π3\frac{2 \pi}{3}) is correct because these are the solutions to the equation derived from the infinite geometric series, within the given interval.
58 Easy
58 of 581
Application +4 | -1
The numerical value of 3cosec⁡20∘−sec⁡20∘\sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ} is
View answer and explanation
Correct option: B
4
The correct answer is 4 because the expression simplifies to 4sin⁡40∘sin⁡40∘\frac{{4 \sin 40^{\circ}}}{{\sin 40^{\circ}}}, which equals 4.
59 Medium
59 of 581
Analyzing +4 | -1
The number of solutions of sin⁡3x=cos⁡2x\sin 3 x=\cos 2 x, in the interval (π2,π)\left(\frac{\pi}{2}, \pi\right) is
View answer and explanation
Correct option: D
1
Option D (1) is correct because after simplifying the equation sin⁡3x=cos⁡2x\sin 3x = \cos 2x and considering the interval (π2,π)(\frac{\pi}{2}, \pi), there is only one solution.
60 Medium
60 of 581
Analyzing +4 | -1
If xx and yy are acute angles such that cos⁡x+cos⁡y=32\cos x+\cos y=\frac{3}{2} and sin⁡x+sin⁡y=34\sin x+\sin y=\frac{3}{4}, then sin⁡(x+y)\sin (x+y) equals
View answer and explanation
Correct option: D
45\frac{4}{5}
Option D (45\frac{4}{5}) is correct because it is the value of sin⁡(x+y)\sin(x+y) calculated using the given equations and trigonometric identities.
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