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JEE • MATHEMATICS • TRIGONOMETRIC EQUATIONS

Trigonometric equations practice · Easy

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Practice format and chapter quiz

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 261 questions
1 Easy
1 of 261
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 261
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 Easy
3 of 261
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.
4 Easy
4 of 261
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}.
5 Easy
5 of 261
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.
6 Easy
6 of 261
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.
7 Easy
7 of 261
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.
8 Easy
8 of 261
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.
9 Easy
9 of 261
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.
10 Easy
10 of 261
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.
11 Easy
11 of 261
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.
12 Easy
12 of 261
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.
13 Easy
13 of 261
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.
14 Easy
14 of 261
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.
15 Easy
15 of 261
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.

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16 Easy
16 of 261
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.
17 Easy
17 of 261
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.
18 Easy
18 of 261
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.
19 Easy
19 of 261
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.
20 Easy
20 of 261
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.
21 Easy
21 of 261
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.
22 Easy
22 of 261
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].
23 Easy
23 of 261
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.
24 Easy
24 of 261
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].
25 Easy
25 of 261
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.
26 Easy
26 of 261
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.
27 Easy
27 of 261
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.
28 Easy
28 of 261
Analyzing +4 | -1
The number of solutions of 2  cos⁡2(x2)sin⁡2  x=x2+1x2,  0≤   x   ≤π22\,\,{\cos ^2}\left( {\frac{x}{2}} \right){\sin ^2}\,\,x = {x^2} + \frac{1}{{{x^2}}},\,\,0 \le \,\,\,x\,\,\, \le \frac{\pi }{2} is
View answer and explanation
Correct option: A
0
Option A (0) is correct because the left-hand side (LHS) of the equation is less than 2, while the right-hand side (RHS) is greater than or equal to 2, implying no solution.
29 Easy
29 of 261
Application +4 | -1
The general solution of the trigonometrical equation sin⁡x+cos⁡x=1\sin x+\cos x=1 is
View answer and explanation
Correct option: C
x=nπ+(−1)nπ/4−π/4,n∈Ix=n \pi+(-1)^n \pi / 4-\pi / 4, n \in \mathbf{I}
Option C (x=nπ\pi+(-1)^n π\pi / 4-π\pi / 4, n ∈\in I) is correct because it represents the general solution derived from transforming the equation into a standard trigonometric form.
30 Easy
30 of 261
Application +4 | -1
If sin⁡θ+cosec⁡θ=2\sin \theta+\operatorname{cosec} \theta=2, then cos⁡2015θ+cosec⁡2015θ\cos ^{2015} \theta+\operatorname{cosec}^{2015} \theta is equal to:
View answer and explanation
Correct option: C
1
Option C (1) is correct because if sin⁡θ+csc⁡θ=2\sin \theta + \csc \theta = 2, then sin⁡θ=1\sin \theta = 1, which implies cos⁡θ=0\cos \theta = 0 and csc⁡θ=1\csc \theta = 1. Therefore, cos⁡2015θ+csc⁡2015θ=0+1=1\cos^{2015} \theta + \csc^{2015} \theta = 0 + 1 = 1.
Revision notes and key takeaways

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Key Takeaways

  • ✓ This set contains 261 single-correct-answer MCQs on TRIGONOMETRIC EQUATIONS.
  • ✓ Try each question before opening its answer. Read the explanation where one is available.
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