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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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Showing 1-30 of 581 questions
1 Easy
1 of 581
Analyzing +4 | -1
The number of solutions of the equation sin⁡  x    cos  3x=sin⁡  3x  cos⁡  5x\sin \,\,x\,\,\;cos\,\,3x = \sin \,\,3x\,\,\cos \,\,5x in [0,   π2]\left[ {0,\;\,\frac{\pi }{2}} \right] is
View answer and explanation
Correct option: C
5
Option C (5) is correct because solving the trigonometric equation leads to 5 distinct solutions within the specified interval [0, π/2].
2 Easy
2 of 581
Analyzing +4 | -1
The number of ordered pairs (x, y)\left( {x,\,y} \right) where x,  y∈[0,10]x,\,\,y \in \left[ {0,10} \right] satisfying (sin⁡2−sin⁡  x+12)2sec⁡2y  ⋅  ≤1\left( {\sqrt {{{\sin }^2} - \sin \,\,x + \frac{1}{2}} } \right){2^{{{\sec }^2}y}}\,\, \cdot \,\, \le 1 is
View answer and explanation
Correct option: B
16
Option B (16) is correct because the given inequality holds only when sin⁡x=12\sin x = \frac{1}{2} and sec⁡2y=1{\sec ^2}y = 1, which gives 4 possible values for x and 4 possible values for y in the given interval, resulting in 16 ordered pairs.
3 Medium
3 of 581
Understanding +4 | -1
Statement I: If slope of the tangent line is zero, then tangent line is perpendicular to the X\mathrm{X}-axis.

Statement II: If θ→π2\theta \rightarrow \frac{\pi}{2}, then tangent line is parallel to the Y-axis.
View answer and explanation
Correct option: B
Only statement II is true
Option B is correct because statement II is true, while statement I is false; a zero slope means the tangent is parallel to the x-axis.
4 Medium
4 of 581
Application +4 | -1
If sin⁡x+cos⁡x=72\sin x+\cos x=\frac{\sqrt{7}}{2}, where x∈[0,π4]x \in\left[0, \frac{\pi}{4}\right], then tan⁡x2\tan \frac{x}{2} is equal to
View answer and explanation
Correct option: B
7−23\frac{\sqrt{7}-2}{3}
Option B (7−23\frac{{\sqrt{{7}}-2}}{{3}}) is correct because it is the solution to the quadratic equation derived from the given trigonometric equation and the constraint on x.
5 Medium
5 of 581
Analyzing +4 | -1
The value of tan⁡π5+2tan⁡2π5+4cot⁡4π5\tan \frac{\pi}{5}+2 \tan \frac{2 \pi}{5}+4 \cot \frac{4 \pi}{5} is
View answer and explanation
Correct option: A
cot⁡π5\cot \frac{\pi}{5}
Option A (cot⁡π5\cot \frac{\pi}{5}) is correct because simplifying the expression using trigonometric identities leads to cot⁡π5\cot \frac{\pi}{5}.
6 Medium
6 of 581
Application +4 | -1
If 0<x<π0 < x < \pi and cos⁡x+sin⁡x=12\cos x+\sin x=\frac{1}{2}, then tan⁡x\tan x is equal to
View answer and explanation
Correct option: B
−(4+7)3-\frac{(4+\sqrt{7})}{3}
Option B (−(4+7)3-\frac{(4+\sqrt{7})}{3}) is correct because it is the value of tan⁡x\tan x calculated using the given equation and the half-angle formula.
7 Easy
7 of 581
Application +4 | -1
If force P=10 kg{P}=10 {~kg} is applied at an angle of 60∘60^{\circ} from the vertical to a block of 20 kg20 {~kg} and the block is in limiting equilibrium, then μ\mu is
View answer and explanation
Correct option: C
34\frac{\sqrt{3}}{4}
Option C (34\frac{\sqrt{3}}{4}) is correct because it is calculated by resolving forces and applying the condition for limiting equilibrium.
8 Medium
8 of 581
Application +4 | -1
General solution of sin⁡  x+cos⁡  x=mina∈R  {1,   a2−4a+6}\sin \,\,x + \cos \,\,x = \begin{array}{cccccccccccccccccccc} {min}\\ {a \in R} \end{array}\,\,\left\{ {1,\,\;{a^2} - 4a + 6} \right\} is

View answer and explanation
Correct option: D
nπ+(−1)n  π4−π4n\pi + {\left( { - 1} \right)^n}\,\,\frac{\pi }{4} - \frac{\pi }{4}
Option D (nπ+(−1)n  π4−π4n\pi + {(- 1)^n}\,\,\frac{\pi }{4} - \frac{\pi }{4}) is correct because it represents the general solution of the given trigonometric equation.
9 Medium
9 of 581
Application +4 | -1
Two sides of a triangle are given by the roots of the equation x2−23x+2=0x^2-2 \sqrt{3} x+2=0. The angle between the sides is π3\frac{\pi}{3}. The perimeter of the triangle is
View answer and explanation
Correct option: B
23+62 \sqrt{3}+\sqrt{6}
Option B (23+62 \sqrt{3}+\sqrt{6}) is correct. By using the given quadratic equation to find the two sides and the given angle, the third side and hence the perimeter can be calculated.
10 Easy
10 of 581
Application +4 | -1
The general value of θ\theta which satisfies the equation (cos⁡θ+isin⁡θ)(cos⁡3θ+isin⁡3θ)…(\cos \theta+i \sin \theta)(\cos 3 \theta+i \sin 3 \theta) \ldots. {cos⁡(2r−1)θ+isin⁡(2r−1)θ}=1\{\cos (2 r-1) \theta+i \sin (2 r-1) \theta\}=1 is
View answer and explanation
Correct option: D
2nπr2\frac{2 n \pi}{r^2}
Option D (2nπr2\frac{2 n \pi}{r^2}) is correct because simplifying the given expression using De Moivre's theorem leads to the general solution θ=2nπr2\theta = \frac{2n\pi}{r^2}.
11 Easy
11 of 581
Application +4 | -1
If 1+sin⁡  θ+sin⁡2  θ+...∞=4+2 3,   0<θ<π,   θ≠π21 + \sin \,\,\theta + {\sin ^2}\,\,\theta + ...\infty = 4 + 2\,\sqrt 3 ,\,\;0 < \theta < \pi ,\;\,\theta \ne \frac{\pi }{2}, then

View answer and explanation
Correct option: D
θ=π3\theta = \frac{\pi }{3} or θ=2π3\theta = \frac{{2\pi }}{3}
The correct answer is D (θ=π3\theta = \frac{\pi }{3} or θ=2π3\theta = \frac{{2\pi }}{3}) because these are the solutions to sin⁡θ=32\sin \theta = \frac{\sqrt{3}}{2} in the given interval.
12 Easy
12 of 581
Application +4 | -1
The value of sin⁡600∘cos⁡330∘+cos⁡120∘sin⁡150∘\sin 600^{\circ} \cos 330^{\circ}+\cos 120^{\circ} \sin 150^{\circ} is
View answer and explanation
Correct option: A
−1-1
Option A (-1) is correct because simplifying the expression using trigonometric identities leads to -1.
13 Medium
13 of 581
Analyzing +4 | -1
If the equation cos⁡4θ+sin⁡2θ+λ=0\cos ^4 \theta+\sin ^2 \theta+\lambda=0 has real solution for θ\theta, then λ\lambda lies in the interval
View answer and explanation
Correct option: A
[−1,−34]\left[-1,-\frac{3}{4}\right]
Option B (32\frac{3}{2}) is correct because after calculating the eccentricities and using the given relationship, the length of the latus rectum of the ellipse can be found to be 3/2.
14 Easy
14 of 581
Understanding +4 | -1
Assertion: Two real numbers or two algebraic expressions related by the symbol <,>,≤<,>, \leq or ≥\geq forms an inequality.

Reason: The inequality ax+by<0ax + by < 0 is strict inequality.
View answer and explanation
Correct option: B
Assertion is correct, reason is correct; reason is not a
correct explanation for assertion
Option A (Only I is true) is correct because the formula for the number of diagonals is correct, but the number of sides for a polygon with 44 diagonals is 11, not 10.
15 Easy
15 of 581
Application +4 | -1
The least positive root of the function sin⁡x−π/2+1=0\sin x-\pi / 2+1=0 lies in the interval
View answer and explanation
Correct option: A
[0,π/2][0, \pi / 2]
Option A ([0,π/2][0, \pi / 2]) is correct because the least positive root of the given function lies in this interval, as sin⁡x=π2−1=47\sin x = \frac{\pi}{2} - 1 = \frac{4}{7} which is between 0 and 1.

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16 Medium
16 of 581
Analyzing +4 | -1
For x∈(0,π)x \in(0, \pi), the equation sin⁡x+2sin⁡2x−sin⁡3x=3\sin x+2 \sin 2 x-\sin 3 x=3 has
View answer and explanation
Correct option: D
No solution
Option D (No solution) is correct because the equation simplifies to a form where sin⁡x\sin x must be 1 and cos⁡x\cos x must be 1/2 simultaneously, which is impossible within the given interval.
17 Medium
17 of 581
Analyzing +4 | -1
If cos⁡2B=cos⁡(A+C)cos⁡(A−C)\cos 2 B=\frac{\cos (A+C)}{\cos (A-C)}, then
View answer and explanation
Correct option: B
tan⁡A,tan⁡B\tan A, \tan B and tan⁡C\tan C are in GP
Option B (tan⁡A,tan⁡B\tan A, \tan B and tan⁡C\tan C are in GP) is correct because by applying componendo and dividendo to the given equation, it can be shown that tan⁡2B=tan⁡Atan⁡C\tan^2 B = \tan A \tan C, which is the condition for a geometric progression.
18 Easy
18 of 581
Application +4 | -1
If e(cos⁡2x+cos⁡4x+cos⁡6x+……∞)log⁡e2e^{\left(\cos ^2 x+\cos ^4 x+\cos ^6 x+\ldots \ldots \infty\right) \log _e^2} satisfies the equation t2−9t+8=0t^2-9 t+8=0, then the value of 2sin⁡xsin⁡x+3cos⁡x(0<x<π2)\frac{{2\sin x}}{{\sin x + \sqrt 3 \cos x}}\left( {0 < x < \frac{\pi }{2}} \right) is
View answer and explanation
Correct option: D
1/21 / 2
Option D (1/2) is correct because simplifying the exponential expression leads to an equation involving cot2(x), which can be solved to find sin(x) and cos(x), and then substituted into the given expression.
19 Easy
19 of 581
Analyzing +4 | -1
The number of solutions of the equation 1+sin⁡  x  sin⁡2  x2=01 + \sin \,\,x\,\,{\sin ^2}\,\,\frac{x}{2} = 0, in [−π,   π]\left[ { - \pi ,\,\;\pi } \right] is

View answer and explanation
Correct option: A
Zero
Option A (Zero) is correct because the equation 1+sin⁡xsin⁡2x2=01 + \sin x \sin^2 \frac{x}{2} = 0 has no solutions in the interval [−π,π][-\pi, \pi] since the expression is always positive.
20 Medium
20 of 581
Application +4 | -1
If sin⁡2θ−2cos⁡θ+14=0\sin ^2 \theta-2 \cos \theta+\frac{1}{4}=0, then the general value of θ\theta is
View answer and explanation
Correct option: B
2nπ±π32 n \pi \pm \frac{\pi}{3}
Option B (2nπ±π32 n \pi \pm \frac{{\pi}}{{3}}) is correct because it represents the general solution derived from solving the given trigonometric equation.
21 Hard
21 of 581
Analyzing +4 | -1
The least positive non-integral solution of sin⁡  π(x2+x)−sin⁡  πx2=0\sin \,\,\pi \left( {{x^2} + x} \right) - \sin \,\,\pi {x^2} = 0, is
View answer and explanation
Correct option: C
Irrational of the formp−14\frac{{\sqrt p - 1}}{4}, when p+14\frac{{\sqrt p + 1}}{4} is an odd integer
Option C is correct because the solution to the given trigonometric equation results in an irrational number of the form (p\sqrt{p} - 1)/4, where (p\sqrt{p} + 1)/4 is an odd integer.
22 Medium
22 of 581
Application +4 | -1
If 1 lies between the roots of 3x2−3(sin⁡)θx−2cos⁡2θ=03 x^2-3(\sin ) \theta x-2 \cos ^2 \theta=0 then complete set of values of sin⁡θ\sin \theta is
View answer and explanation
Correct option: C
12<sin⁡θ<1\frac{1}{2} < \sin \theta < 1
Option C (12<sin⁡θ<1\frac{{1}}{{2}} < \sin \theta < 1) is correct because when 1 lies between the roots, f(1) < 0, which leads to the given range for sin⁡θ\sin \theta.
23 Medium
23 of 581
Application +4 | -1
2sin⁡2x+sin⁡22x=2,−π<x<π2 \sin ^{2} x+\sin ^{2} 2 x=2,-\pi < x < \pi, then x=x=
View answer and explanation
Correct option: B
±π4\pm \frac{\pi}{4}
Option B (±π4\pm \frac{{\pi}}{{4}}) is correct because it is a solution to the trigonometric equation within the given interval.
24 Easy
24 of 581
Application +4 | -1
If tan⁡    3θ−1tan⁡    3θ+1=3\frac{{\tan \;\,\,3\theta - 1}}{{\tan \;\,\,3\theta + 1}} = \sqrt 3, then the general value of θ\theta is
View answer and explanation
Correct option: C
nπ3+7π36\frac{{n\pi }}{3} + \frac{{7\pi }}{{36}}
Option C (nπ3+7π36\frac{{n\pi }}{3} + \frac{{7\pi }}{{36}}) is correct because it represents the general solution for θ\theta after solving the given trigonometric equation.
25 Easy
25 of 581
Analyzing +4 | -1
The solution of the equation sin⁡10  2x=1+cos⁡10  x{\sin ^{10}}\,\,2x = 1 + {\cos ^{10}}\,\,x is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the given equation has no solution as the maximum value of LHS is 1 and the minimum value of RHS is 1, but they cannot be simultaneously satisfied.
26 Medium
26 of 581
Application +4 | -1
The sum of the solutions in (0,2π)(0,2 \pi) of the equation cos⁡xcos⁡(π3−x)cos⁡(π3+x)=14\cos x \cos \left(\frac{\pi}{3}-x\right) \cos \left(\frac{\pi}{3}+x\right)=\frac{1}{4} is
View answer and explanation
Correct option: C
2π2 \pi
Option C (2π2 \pi) is correct because using the trigonometric identity, the sum of solutions in (0, 2π) is 2π2\pi.
27 Easy
27 of 581
Application +4 | -1
If cos⁡  θ=−32\cos \,\,\theta = - \frac{{\sqrt 3 }}{2} and sin⁡  α=−35\sin \,\,\alpha = - \frac{3}{5}, where θ\theta does not lie in the third quadrant, then 2tan⁡α+3tan⁡θcot⁡2θ+cos⁡α\frac{{2\tan \alpha + \sqrt 3 \tan \theta }}{{{{\cot }^2}\theta + \cos \alpha }} is equal to
View answer and explanation
Correct option: B
522\frac{5}{{22}}
Option B (522\frac{5}{{22}}) is correct because substituting the values of tan⁡α\tan \alpha, tan⁡θ\tan \theta, cot⁡θ\cot \theta, and cos⁡α\cos \alpha into the expression and simplifying yields this result.
28 Medium
28 of 581
Application +4 | -1
The ordered pair (x,y)(x, y) satisfying 1−tan⁡x1+tan⁡x=tan⁡y\frac{1-\tan x}{1+\tan x}=\tan y and x−y=π6x-y=\frac{\pi}{6} is
View answer and explanation
Correct option: A
(nπ2+π24,nπ2+π24),∈Z\left(\frac{n \pi}{2}+\frac{\pi}{24}, \frac{n \pi}{2}+\frac{\pi}{24}\right), \quad \in \mathbb{Z}
Option C ((nπ2\frac{{nπ}}{{2}} + 5π24\frac{{5π}}{{24}}, nπ2\frac{{nπ}}{{2}} - π24\frac{{π}}{{24}}), n ∈ \mathbb{{Z}}) is correct because it represents the general solution for x and y derived from the given equations using trigonometric identities.
29 Medium
29 of 581
Analyzing +4 | -1
If ntan⁡y=tan⁡x,n∈R+n \tan y=\tan x, n \in R^{+}, then the maximum value of sec⁡2(x−y)\sec ^2(x-y) is equal to
View answer and explanation
Correct option: D
(n+1)24n\frac{(n+1)^2}{4 n}
Option A (nπ−π4n \pi-\frac{{\pi}}{{4}}) is correct because it represents the general solution where the real part of the given complex expression is zero, making it purely imaginary.
30 Medium
30 of 581
Application +4 | -1
A solution (x,y)(x, y) of the system of equation x−y=1/3x-y=1 / 3 and cos⁡2(πx)−sin⁡2(πy)=1/2\cos ^2(\pi x)-\sin ^2(\pi y)=1 / 2 is given by
View answer and explanation
Correct option: C
(13/6,11/6)(13 / 6,11 / 6)
Option C ((13/6,11/6)(13 / 6,11 / 6)) is correct because it satisfies both equations of the given system when n=2n=2.
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