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JEE • MATHEMATICS • TRIGONOMETRIC EQUATIONS

Trigonometric equations practice · Mixed difficulty

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Showing 61-90 of 581 questions

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61 Hard
61 of 581
Application +4 | -1
Solution of the equation sin⁡x+cos⁡x−22sin⁡xcos⁡x=0\sin x+\cos x-2 \sqrt{2} \sin x \cos x=0 is
View answer and explanation
Correct option: A
x=(8k+3)⋅π12x=(8 k+3) \cdot \frac{\pi}{12} or x=2kπ+π4;k∈Zx=2 k \pi+\frac{\pi}{4} ; k \in \mathbb{Z}
Option A is the correct solution as it represents the general solutions derived from solving the trigonometric equation by converting it into sin⁡(x+π4)=sin⁡(2x)\sin(x + \frac{\pi}{4}) = \sin(2x).
62 Medium
62 of 581
Analyzing +4 | -1
If asin⁡x+bcos⁡(x+θ)+bcos⁡(x−θ)=da \sin x+b \cos (x+\theta)+b \cos (x-\theta)=d, then the minimum value of ∣cos⁡θ∣|\cos \theta| is
View answer and explanation
Correct option: A
12∣b∣d2−a2\frac{1}{2|b|} \sqrt{d^2-a^2}
Option A (12∣b∣d2−a2\frac{1}{2|b|} \sqrt{d^2-a^2}) is correct because it represents the minimum value of |cos θ| derived from the given equation.
63 Easy
63 of 581
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.
64 Easy
64 of 581
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.
65 Medium
65 of 581
Understanding +4 | -1
Solution of the inequality sin⁡x>1/2\sin x > 1 / \sqrt{2} belongs to
View answer and explanation
Correct option: A
(2nπ+π4,2nπ+3π4)\left(2 n \pi+\frac{\pi}{4}, 2 n \pi+\frac{3 \pi}{4}\right) where n∈Zn \in \mathbb{Z}
Option A ((2nπ+π4,2nπ+3π4)\left(2 n \pi+\frac{\pi}{4}, 2 n \pi+\frac{3 \pi}{4}\right) where n∈Zn \in \mathbb{{Z}}) is correct because sin⁡x>1/2\sin x > 1/\sqrt{{2}} when xx is between π/4\pi/4 and 3π/43\pi/4 in each period.
66 Easy
66 of 581
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.
67 Medium
67 of 581
Analyzing +4 | -1
The number of solutions of 2sin⁡x+cos⁡x=32 \sin x+\cos x=3 is
View answer and explanation
Correct option: D
No solution
Option D (No solution) is correct because the maximum value of 2sin⁡x+cos⁡x2 \sin x + \cos x is 5\sqrt{5}, which is less than 3.
68 Easy
68 of 581
Application +4 | -1
Solution of the system of equations x+y=π4,tan⁡x+tan⁡y=1x+y=\frac{\pi}{4}, \tan x+\tan y=1 is
View answer and explanation
Correct option: B
x=π4−nπ,y=nπx=\frac{\pi}{4}-n \pi, y=n \pi
Option B (x=π4−nπ,y=nπx=\frac{{\pi}}{{4}}-n \pi, y=n \pi) is the correct solution because it satisfies both given equations.
69 Medium
69 of 581
Analyzing +4 | -1
The total number of ordered pairs (r,  θ)\left( {r,\,\,\theta } \right) satisfying r  sin⁡  θ=3,  r=4(1+sin⁡  θ)r\,\,\sin \,\,\theta = 3,\,\,r = 4\left( {1 + \sin \,\,\theta } \right), where r>0r > 0 and θ∈[−π,  π]\theta \in \left[ { - \pi ,\,\,\pi } \right] is
View answer and explanation
Correct option: B
2\mathrm{2}
Option B (2) is correct because solving the system of equations yields two valid solutions for θ\theta within the given interval.
70 Medium
70 of 581
Analyzing +4 | -1
The number of roots of the equation ∑k=17sin⁡(kx)=7sin⁡4x\sum\limits_{{k=1}}^7 \sin (k x)=7 \sin 4 x in the interval [0,2π][0,2 \pi] is
View answer and explanation
Correct option: D
9
The correct answer is D (9) because the equation simplifies to sin(4x) = 0 or cos(x) = 1, leading to 9 solutions in the interval [0, 2π].
71 Medium
71 of 581
Analyzing +4 | -1
If x1x_1 and x2x_2 are two positive values of xx for which 2cos⁡x,∣cos⁡x∣2 \cos x,|\cos x| and 3sin⁡2x−23 \sin ^2 x-2 are in GP. The minimum value of ∣x1−x2∣\left|x_1-x_2\right| is
View answer and explanation
Correct option: C
2cos⁡−1(23)2 \cos ^{-1}\left(\frac{2}{3}\right)
Option C (2cos⁡−1(23)2 \cos ^{-1}\left(\frac{2}{3}\right)) is correct because it represents the minimum difference between two positive values of x that satisfy the given geometric progression condition.
72 Medium
72 of 581
Application +4 | -1
The general solution of the equation tan⁡θ+tan⁡2θ+3tan⁡θtan⁡2θ=3\tan \theta+\tan 2 \theta+\sqrt{3} \tan \theta \tan 2 \theta=\sqrt{3} is
View answer and explanation
Correct option: A
θ=(3n+1)π9\theta=(3 n+1) \frac{\pi}{9}
Option A (θ=(3n+1)π9\theta=(3 n+1) \frac{{\pi}}{{9}}) is correct because it represents the general solution derived from solving the given trigonometric equation.
73 Medium
73 of 581
Application +4 | -1
If 3sin⁡α=5sin⁡β3 \sin \alpha=5 \sin \beta, then tan⁡α+β2tan⁡α−β2\frac{\tan \frac{\alpha+\beta}{2}}{\tan \frac{\alpha-\beta}{2}} is equal to
View answer and explanation
Correct option: D
4
Option D (4) is correct because by applying the componendo and dividendo rule and the sum-to-product trigonometric identities, the expression simplifies to 4.
74 Easy
74 of 581
Analyzing +4 | -1
The number of real solutions of 2  sin⁡(ex)=5x+5−x2\,\,\sin \left( {{e^x}} \right) = {5^x} + {5^{ - x}} in [0,1] is/ are\left[ {\mathrm{0, 1}} \right]\mathrm{ }\,\mathrm{is /}\,\mathrm{are}
View answer and explanation
Correct option: A
0\mathrm{0}
Option A (0) is correct because the equation 2sin⁡(ex)=5x+5−x2\sin(e^x) = 5^x + 5^{-x} has no real solutions due to the range of the functions involved.
75 Medium
75 of 581
Application +4 | -1
The most general solution of the equation
8  tan⁡2  θ2=1+sec⁡  θ,    is8\,\,{\tan ^2}\,\,\frac{\theta }{2} = 1 + \sec \,\,\theta ,\;\,\,\mathrm{is}
View answer and explanation
Correct option: A
θ=2n  π  ±  cos⁡−1  (13)\theta = 2n\;\pi \,\, \pm \,\,{\cos ^{ - 1}}\,\,\left( {\frac{1}{3}} \right)
Option A (θ=2n  π ± cos⁡−1 (13)\theta = 2n\;\pi \, \pm \,{\cos ^{ - 1}}\,\left( {\frac{1}{3}} \right)) is correct because it represents the general solution obtained by solving the given trigonometric equation.

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76 Medium
76 of 581
Application +4 | -1
The equation kcos⁡x−3sin⁡x=k+1k \cos x-3 \sin x=k+1 is solvable only if
View answer and explanation
Correct option: B
k∈(−∞,4]k \in(-\infty, 4]
Option B (k∈(−∞,4]k \in(-\infty, 4]) is correct because the equation is solvable only when the absolute value of (k+1) is less than or equal to the square root of (k2 + 9), leading to k <= 4.
77 Easy
77 of 581
Application +4 | -1
The general solution of tan⁡3  x=1\tan 3\;x = 1 is
View answer and explanation
Correct option: B
n  π3+π12\frac{{n\;\pi }}{3} + \frac{\pi }{{12}}
Option B (n  π3+π12\frac{{n\;\pi }}{3} + \frac{\pi }{{12}}) is correct because it represents the general solution of the given trigonometric equation.
78 Easy
78 of 581
Analyzing +4 | -1
The number of solutions of sin⁡7x+cos⁡7x=1,  x∈[0, 4π]{\sin ^7}x + {\cos ^7}x = 1,\,\,x \in [0,\,4\pi ] is equal to
View answer and explanation
Correct option: C
5
Option C (5) is correct because the equation sin⁡7x+cos⁡7x=1{\sin ^7}x + {\cos ^7}x = 1 is satisfied only when sin⁡x=0\sin x = 0 or cos⁡x=0\cos x = 0, or sin⁡x=1\sin x = 1 or cos⁡x=1\cos x = 1, leading to 5 solutions in the interval [0,4π][0, 4\pi].
79 Easy
79 of 581
Application +4 | -1
If sin⁡  4A−cos⁡  2A=cos⁡  4A−sin⁡  2A,   (0<A<π4)\sin \,\,4A - \cos \,\,2A = \cos \,\,4A - \sin \,\,2A,\;\,\left( {0 < A < \frac{\pi }{4}} \right), then the value of tan⁡  4A\tan \,\,4A is

View answer and explanation
Correct option: C
3\sqrt 3
Option C (3\sqrt 3) is correct because it is the value of tan⁡4A\tan 4A after solving the given trigonometric equation and finding the value of A.
80 Medium
80 of 581
Analyzing +4 | -1
If sum of all the solutions of the equation 8cos⁡x(cos⁡(π6+x)cos⁡(π6−x)−12)=18 \cos x\left(\cos \left(\frac{\pi}{6}+x\right) \cos \left(\frac{\pi}{6}-x\right)-\frac{1}{2}\right)=1 in [0,π][0, \pi] is kπ{k} \pi, then k{k} is equal to:
View answer and explanation
Correct option: C
139\frac{13}{9}
The value of k is 13/9. This is found by simplifying the trigonometric equation, solving for x, and summing the solutions in the given interval.
81 Easy
81 of 581
Analyzing +4 | -1
If the solution of the equation log⁡cos⁡xcot⁡x+4log⁡sin⁡x\log _{\cos x} \cot x+4 \log _{\sin x} tan⁡x=1,x∈(0,π2)\tan x=1, x \in\left(0, \frac{\pi}{2}\right), is sin⁡−1(α+β2)\sin ^{-1}\left(\frac{\alpha+\sqrt{\beta}}{2}\right), where α,β\alpha, \beta are integers, then α+β\alpha+\beta is equal to
View answer and explanation
Correct option: C
4
Option C (4) is correct because solving the logarithmic equation leads to a quadratic in sin(x), and the sum of the coefficients in the solution is 4.
82 Medium
82 of 581
Analyzing +4 | -1
If cos⁡A=mcos⁡B\cos A=m \cos B and cot⁡(A+B2)=λ/tan⁡(B−A2)\cot \left(\frac{A+B}{2}\right)=\lambda/\tan \left(\frac{B-A}{2}\right), then λ\lambda is equal to
View answer and explanation
Correct option: C
m+1m−1\frac{m+1}{m-1}
Option C (m+1m−1\frac{{m+1}}{{m-1}}) is correct because it's derived by applying the componendo and dividendo rule and trigonometric identities to the given equation.
83 Medium
83 of 581
Understanding +4 | -1
If sin⁡θ=12\sin \theta=\frac{1}{2} and tan⁡θ=13,∀ n∈I\tan \theta=\frac{1}{\sqrt{3}}, \forall \, n \in I, then most general value of θ\theta is
View answer and explanation
Correct option: A
2nπ+π6,∀ n∈I2 n \pi+\frac{\pi}{6}, \forall \, n \in I
The correct answer is 2nπ+π6,∀,n∈I2 n \pi+\frac{{\pi}}{{6}}, \forall \\, n \in I because it represents the general solution that satisfies both sin⁡θ=12\sin \theta = \frac{{1}}{{2}} and tan⁡θ=13\tan \theta = \frac{{1}}{{\sqrt{{3}}}} .
84 Medium
84 of 581
Analyzing +4 | -1
Number of ordered pairs (a,x)(a, x) satisfying the equation sec⁡2(a+2)x+a2−1=0,−π<x<π\sec ^2(a+2) x+a^2-1=0,-\pi < x < \pi is
View answer and explanation
Correct option: C
3
Option C (3) is correct because solving the given equation leads to three ordered pairs that satisfy the conditions.
85 Easy
85 of 581
Application +4 | -1
The most general value of θ\mathrm{\theta } satisfying the equation (1+2sin⁡θ)2+  (3tan⁡θ−1)2=0{(1 + 2\sin \mathrm{\theta })^2} + \,\,{(\sqrt 3 \tan \mathrm{\theta } - 1)^2} = 0 are given by
View answer and explanation
Correct option: C
2nπ+7π62n\pi + \frac{{7\pi }}{6}
Option C (2nπ+7π62n\pi + \frac{{7\pi }}{6}) is correct because it satisfies both equations derived from the given condition.
86 Easy
86 of 581
Application +4 | -1
The set of values of xx for which tan⁡3x−tan⁡2x1+tan⁡3xtan⁡2x=1\frac{\tan 3 x-\tan 2 x}{1+\tan 3 x \tan 2 x}=1 is
View answer and explanation
Correct option: A
ϕ\phi
Option A (ϕ\phi) is correct because the solution obtained from the simplified equation makes the original equation undefined.
87 Medium
87 of 581
Analyzing +4 | -1
If S=cos⁡2πn+cos⁡22πn+…+cos⁡2(n−1)πnS=\cos ^2 \frac{\pi}{n}+\cos ^2 \frac{2 \pi}{n}+\ldots+\cos ^2 \frac{(n-1) \pi}{n}, then SS equals
View answer and explanation
Correct option: C
12(n−2)\frac{1}{2}(n-2)
Option C (12(n−2)\frac{1}{2}(n-2)) is correct because the sum simplifies to this expression using trigonometric identities.
88 Hard
88 of 581
Analyzing +4 | -1
The value of ∑n=1∞tan⁡(θ2n)2n−1cos⁡(θ2n−1)\sum\limits_{{n=1}}^{\infty} \frac{\tan \left(\frac{\theta}{2^n}\right)}{2^{n-1} \cos \left(\frac{\theta}{2^{n-1}}\right)} is
View answer and explanation
Correct option: A
2sin⁡2θ−1θ\frac{2}{\sin 2 \theta}-\frac{1}{\theta}
Option A (2sin⁡2θ−1θ\frac{{2}}{{\sin 2 \theta}} - \frac{{1}}{{\theta}}) is correct because the infinite sum telescopes and simplifies to this expression.
89 Easy
89 of 581
Analyzing +4 | -1
The number of values of xx for which sin⁡  2x+cos⁡  4x=2\sin \,\,2x + \cos \,\,4x = 2, is
View answer and explanation
Correct option: A
0
Option A (0) is correct because the maximum value of both sin(2x) and cos(4x) is 1, and they cannot simultaneously be 1 to satisfy the equation sin(2x) + cos(4x) = 2.
90 Medium
90 of 581
Analyzing +4 | -1
If 2sin⁡α1+cos⁡α+sin⁡α=x\frac{2 \sin \alpha}{1+\cos \alpha+\sin \alpha}=x, then 1−cos⁡α−sin⁡αcos⁡α\frac{1-\cos \alpha-\sin \alpha}{\cos \alpha} is equal to
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because simplifying the given expression leads to 1−cos⁡α−sin⁡αcos⁡α=−x\frac{{1 - \cos \alpha - \sin \alpha}}{{\cos \alpha}} = -x, which is not among the provided options.
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