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JEE • MATHEMATICS • TRIGONOMETRIC EQUATIONS

Trigonometric equations practice · Medium

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Showing 91-120 of 294 questions

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91 Medium
91 of 294
Application +4 | -1
If 4cos⁡2xsin⁡x−2sin⁡2x=3sin⁡x4 \cos ^2 x \sin x-2 \sin ^2 x=3 \sin x, then xx is equal to
View answer and explanation
Correct option: B
nπ+(−1)n+1(3π10)n \pi+(-1)^{n+1}\left(\frac{3 \pi}{10}\right)
Option B (nπ+(−1)n+1(3π10)n \pi+(-1)^{n+1}(\frac{3 \pi}{10})) is correct because by solving the given trigonometric equation, we find the general solution for x.
92 Medium
92 of 294
Application +4 | -1
If cos⁡x−cot⁡βsin⁡x2=32\cos x-\frac{\cot \beta \sin x}{2}=\frac{\sqrt{3}}{2}, then the value of tan⁡x2\tan \frac{x}{2} is
View answer and explanation
Correct option: A
tan⁡β2tan⁡15∘\tan \frac{\beta}{2} \tan 15^{\circ}
Option A (tan⁡β2tan⁡15∘\tan \frac{\beta}{2} \tan 15^{\circ}) is correct because it is derived by solving the given trigonometric equation for tan⁡x2\tan \frac{x}{2}.
93 Medium
93 of 294
Application +4 | -1
Let fk(x)=1k(sin⁡kx+cos⁡kx)f_k(x)=\frac{1}{k}\left(\sin ^k x+\cos ^k x\right), where, x∈Rx \in R and k≥1k \geq 1. Then, f4(x)−f6(x)f_4(x)-f_6(x) equals
View answer and explanation
Correct option: D
112\frac{1}{12}
Option D (112\frac{1}{12}) is correct because after simplifying the expression f4(x)−f6(x)f_4(x) - f_6(x), it equals 112\frac{1}{12}.
94 Medium
94 of 294
Application +4 | -1
If max⁡x  ∈  R   {5  sin⁡  x+3  sin⁡  (x−θ)}=7\mathop {\max }\limits_{x\; \in \;R} \,\;\left\{ {5\,\,\sin \,\,x + 3\,\,\sin \,\,\left( {x - \theta } \right)} \right\} = 7, then θ=\theta =
View answer and explanation
Correct option: A
2  n  π  ±  π3,  n∈Z2\;n\;\pi \,\, \pm \,\,\frac{\pi }{3},\,\,n \in Z
Option A (2  n  π;±;π3,;n∈Z2\;n\;\pi \\; \pm \\; \frac{\pi }{{3}},\\;n \in Z) is correct because it satisfies the condition where the maximum value of the expression is 7.
95 Medium
95 of 294
Analyzing +4 | -1
If sin⁡  6θ=32  cos⁡5  θ  sin⁡  θ−32  cos⁡3θ  sin⁡  θ+3x\sin \,\,6\theta = 32\,\,{\cos ^5}\,\,\theta \,\,\sin \,\,\theta - 32\,\,{\cos ^3}\theta \,\,\sin \,\,\theta + 3x, then xx is equal to
View answer and explanation
Correct option: D
sin⁡  2θ\sin \,\,2\theta
Option D (sin⁡ 2θ\sin \,2\theta) is correct because by using the triple angle formula for sine and simplifying, we find that x=sin⁡ 2θx = \sin \,2\theta.
96 Medium
96 of 294
Application +4 | -1
If 2sin⁡θ−1=02 \sin \theta-1=0 and 2cos⁡θ+3=2 \cos \theta+\sqrt{3}= 0 then genrate value of θ\theta are given by
View answer and explanation
Correct option: D
2nπ+5π62 \mathrm{n} \pi+\frac{5 \pi}{6}
Option D (2nπ+5π62 \mathrm{n} \pi+\frac{5 \pi}{6}) is correct because it represents the general solution for θ that satisfies both sin⁡θ=12\sin \theta = \frac{1}{2} and cos⁡θ=−32\cos \theta = -\frac{\sqrt{3}}{2}, placing θ in the second quadrant.
97 Medium
97 of 294
Application +4 | -1
If sin⁡  3θ=sin⁡  θ\sin \,\,3\theta = \sin \,\,\theta, how many solutions exist such that 0<θ<2π0 < \theta < 2\pi ?
View answer and explanation
Correct option: C
5\mathrm{5}
Option C (5) is the correct answer because solving the trigonometric equation sin⁡3θ=sin⁡θ\sin 3\theta = \sin \theta within the given interval yields five distinct solutions.
98 Medium
98 of 294
Application +4 | -1
If sin⁡θ=2t1+t2\sin \theta=\frac{2 t}{1+t^2} and θ\theta lies in second quadrant, then cos⁡θ\cos \theta is equal to
View answer and explanation
Correct option: C
−∣1−t2∣1+t2\frac{-\left|1-t^2\right|}{1+t^2}
Option C (−∣1−t2∣1+t2\frac{{-\left|1-t^2\right|}}{{1+t^2}}) is correct because, given sin⁡θ=2t1+t2\sin \theta = \frac{{2t}}{{1+t^2}} and θ\theta is in the second quadrant, cos⁡θ\cos \theta must be negative, leading to the expression −∣1−t2∣1+t2\frac{{-\left|1-t^2\right|}}{{1+t^2}}.
99 Medium
99 of 294
Analyzing +4 | -1
Let f(x)=x2f(x)=x^2 and g(x)=sin⁡xg(x)=\sin x, for all x∈Rx \in R. Then, the set of all xx satisfying (fogogof)(x)=(gogof)(x)(fogogof) (x)=({gog} o f)(x), where (fog)(x)=f[g(x)](fog)(x)=f[g(x)], is
View answer and explanation
Correct option: A
±nπ,n∈{0,1,2,…}\pm \sqrt{n \pi}, n \in\{0,1,2, \ldots\}
Option A (±nπ,n∈0,1,2,…\pm \sqrt{n \pi}, n \in{{0,1,2, \ldots}}) is correct because it represents the set of all x values that satisfy the given functional equation after simplification and solving for x.
100 Medium
100 of 294
Application +4 | -1
Let I(n)=2cos⁡nx,∀ n∈NI(n)=2 \cos n x, \forall \, n \in N, then I(1)⋅I(n+1)−I(n)I(1) \cdot I(n+1)-I(n) is equal to
View answer and explanation
Correct option: B
I(n+2)I(n+2)
Option B (I(n+2)I(n+2)) is correct because by using the trigonometric identity 2cos⁡acos⁡b=cos⁡(a+b)+cos⁡(a−b)2 \cos a \cos b = \cos(a+b) + \cos(a-b), the expression simplifies to I(n+2)I(n+2).
101 Medium
101 of 294
Analyzing +4 | -1
The different between the greatest and least values of the function f(x)=cos⁡x12cos⁡2x−13cos⁡3x  f\left( x \right) = \cos x\frac{1}{2}\cos 2x - \frac{1}{3}\cos 3x\; is
View answer and explanation
Correct option: D
94\frac{9}{4}
Option D (94\frac{9}{4}) is correct because it represents the difference between the greatest and least values of the function, calculated by finding critical points and evaluating the function at those points.
102 Medium
102 of 294
Understanding +4 | -1
The equation sin⁡x+sin⁡y+sin⁡z=−3\sin x+\sin y+\sin z=-3 for 0≤x≤2π,0≤y≤2π,0≤z≤2π0 \leq x \leq 2 \pi, 0 \leq y \leq 2 \pi, 0 \leq z \leq 2 \pi, has
View answer and explanation
Correct option: A
One solution
Option A (One solution) is correct because the equation is only satisfied when each sine term is -1, leading to a unique solution.
103 Medium
103 of 294
Analyzing +4 | -1
Let S={0∈(0,π2):∑m=19sec⁡(θ+(m−1)π6)sec⁡(θ+mπ6)=−83}.S = \left\{ {0 \in \left( {0,\frac{\pi }{2}} \right):\sum\limits_{m = 1}^9 {\sec \left( {\theta + \left( {m - 1} \right)\frac{\pi }{6}} \right)\sec \left( {\theta + \frac{{m\pi}}{6}} \right) = - \frac{8}{{\sqrt 3 }}} } \right\}. Then
View answer and explanation
Correct option: C
∑θ∈Sθ=π2\sum\limits_{\theta \in S} {\theta = \frac{\pi }{2}}
Option C (∑θ∈Sθ=π2\sum\limits_{\theta \in S} {\theta = \frac{\pi }{2}}) is correct because solving the trigonometric equation leads to two solutions for θ\theta, and their sum is π2\frac{\pi}{2}.
104 Medium
104 of 294
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
View answer and explanation
Correct option: C
5
Option C (5) is correct because we need to find the values of x in the interval [0,4π][0, 4\pi] for which sin⁡7x+cos⁡7x=1\sin^7 x + \cos^7 x = 1, which occurs when either sin⁡x=0\sin x = 0 and cos⁡x=1\cos x = 1 or sin⁡x=1\sin x = 1 and cos⁡x=0\cos x = 0.
105 Medium
105 of 294
Analyzing +4 | -1
If sin⁡  θ−cos⁡  θ<0\sin \,\,\theta - \cos \,\,\theta < 0, then θ\theta lies between
View answer and explanation
Correct option: D
2n  π−3π42n\;\pi - \frac{{3\pi }}{4} and 2n  π+π4,   n∈Z2n\;\pi + \frac{\pi }{4},\;\,n \in Z
Option A is correct because it represents the range of θ where sin(θ) - cos(θ) < 0, derived by transforming the inequality and solving for θ.

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SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs
106 Medium
106 of 294
Application +4 | -1
If 0≤θ≤π20 \leq \theta \leq \frac{\pi}{2} and x=Xcos⁡θ+Ysin⁡θx=X \cos \theta+Y \sin \theta, y=Xsin⁡θ−Ycos⁡θy=X \sin \theta-Y \cos \theta such that x2+2xy+y2=aX2+bY2x^2+2 x y+y^2=a X^2+b Y^2, where aa and bb are constants, then
View answer and explanation
Correct option: C
a=3,b=−1a=3, b=-1
Option C (a=3,b=−1a=3, b=-1) is correct because by substituting the given expressions for x and y into the equation and comparing coefficients, we find that a=3a=3 and b=−1b=-1 when θ=π4\theta = \frac{{\pi}}{{4}}.
107 Medium
107 of 294
Analyzing +4 | -1
The value of 2cos⁡π13cos⁡9π13+cos⁡3π13+cos⁡5π132 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13} is
View answer and explanation
Correct option: B
0
Option B (0) is correct because the expression simplifies to zero using trigonometric identities.
108 Medium
108 of 294
Analyzing +4 | -1
If α\alpha and β\beta satisfy the equation 12sin⁡α+5cos⁡α12 \sin \alpha+5 \cos \alpha =2β2−8β+21=2 \beta^2-8 \beta+21, then the value of β2−2β\beta^2-2 \beta is _____.
View answer and explanation
Correct option: A
0
Option A (0) is correct because the maximum value of 12sin⁡α+5cos⁡α12 \sin \alpha + 5 \cos \alpha is 13, and the minimum value of 2β2−8β+212 \beta^2 - 8 \beta + 21 is also 13, which occurs when β=2\beta = 2. Therefore, β2−2β=0\beta^2 - 2\beta = 0.
109 Medium
109 of 294
Application +4 | -1
If 2sec⁡θ+tan⁡θ=1\sqrt{2} \sec \theta+\tan \theta=1, then the general value θ\theta is
View answer and explanation
Correct option: C
2nπ−π42 n \pi-\frac{\pi}{4}
Option C (2nπ−π42 n \pi-\frac{{\pi}}{{4}}) is correct because it represents the general solution derived from solving the given trigonometric equation.
110 Medium
110 of 294
Analyzing +4 | -1
The value of the expression sin⁡3x1+cos⁡x+cos⁡3x1−sin⁡x\frac{\sin ^3 x}{1+\cos x}+\frac{\cos ^3 x}{1-\sin x} is
View answer and explanation
Correct option: A
2cos⁡[π4−x]\sqrt{2} \cos \left[\frac{\pi}{4}-x\right]
Option A (2cos⁡[π4−x]\sqrt{2} \cos \left[\frac{\pi}{4}-x\right]) is correct because the expression simplifies to sin⁡x+cos⁡x\sin x + \cos x, which is equivalent to 2cos⁡[π4−x]\sqrt{2} \cos \left[\frac{\pi}{4}-x\right].
111 Medium
111 of 294
Application +4 | -1
If 2  sin⁡2θ=3  cos⁡  θ,  0≤θ≤2π2\,\,{\sin ^2}\mathrm{\theta } = 3\,\,\cos \,\,\mathrm{\theta },\mathrm{\;}0 \le \mathrm{\theta } \le 2\mathrm{\pi } then θ\mathrm{\theta } is equal to

View answer and explanation
Correct option: C
π3,  5 π3\frac{\mathrm{\pi }}{3},\,\,\frac{{5\,\mathrm{\pi }}}{3}
Option C (π3,  5 π3\frac{\mathrm{\pi }}{3},\,\,\frac{{5\,\mathrm{\pi }}}{3}) is correct because these values of θ\theta satisfy the given trigonometric equation within the specified range of 0≤θ≤2π0 \le \mathrm{\theta } \le 2\mathrm{\pi }.
112 Medium
112 of 294
Analyzing +4 | -1
If cos⁡α+2cos⁡β+3cos⁡γ=0,sin⁡α+2sin⁡β\cos \alpha+2 \cos \beta+3 \cos \gamma=0, \sin \alpha+2 \sin \beta +3sin⁡γ=0+3 \sin \gamma=0 and α+β+γ=π\alpha+\beta+\gamma=\pi, then sin⁡3α+8sin⁡3β+27sin⁡3γ\sin 3 \alpha+8 \sin 3 \beta+27 \sin 3 \gamma is equal to
View answer and explanation
Correct option: B
0
Option B is correct because using complex numbers and the given conditions, it can be shown that sin⁡3α+8sin⁡3β+27sin⁡3γ=0\sin 3 \alpha+8 \sin 3 \beta+27 \sin 3 \gamma = 0.
113 Medium
113 of 294
Analyzing +4 | -1
The number of solutions of the equation (1−cos⁡2x)(1-\cos 2 x) (cos⁡2x+cos⁡2x)=0,0≤x≤2π\left(\cos 2 x+\cos ^2 x\right)=0,0 \leq x \leq 2 \pi is
View answer and explanation
Correct option: D
0
Option D (0) is correct because the given equation has no solutions in the specified interval due to the cotangent function being undefined.
114 Medium
114 of 294
Application +4 | -1
If tan⁡α=ba,a>b>0\tan \alpha=\frac{b}{a}, a > b > 0 and 0<α<π40 < \alpha < \frac{\pi}{4}, then a+ba−b−a−ba+b\sqrt{\frac{a+b}{a-b}}-\sqrt{\frac{a-b}{a+b}} is equal to
View answer and explanation
Correct option: A
2sin⁡αcos⁡2α\frac{2 \sin \alpha}{\sqrt{\cos 2 \alpha}}
Option A (2sin⁡αcos⁡2α\frac{{2 \sin \alpha}}{{\sqrt{{\cos 2 \alpha}}}}) is correct because it is derived by simplifying the given expression using trigonometric identities and the given condition tan⁡α=ba\tan \alpha = \frac{{b}}{{a}}.
115 Medium
115 of 294
Application +4 | -1
The most general solutions of the equation sec⁡2x=2(1−tan⁡2x)\sec ^2 x=\sqrt{2}\left(1-\tan ^2 x\right) are given by
View answer and explanation
Correct option: C
nπ±π8n \pi \pm \frac{\pi}{8}
Option C (nπ±π8n \pi \pm \frac{\pi}{8}) is correct because by simplifying the given equation, we get tan⁡2x=tan⁡2π8\tan^2 x = \tan^2 \frac{\pi}{8}, which leads to the general solution x=nπ±π8x = n \pi \pm \frac{\pi}{8}.
116 Medium
116 of 294
Application +4 | -1
If sin⁡α,1\sin \alpha, 1 and cos⁡2α\cos 2 \alpha are in GP, then α\alpha is equal to
View answer and explanation
Correct option: B
nπ+(−1)n−1π2,n∈In \pi+(-1)^{n-1} \frac{\pi}{2}, n \in I
Option B (nπ+(−1)n−1π2,n∈In \pi+(-1)^{n-1} \frac{\pi}{2}, n \in I) is correct because when sin⁡α,1\sin \alpha, 1 and cos⁡2α\cos 2 \alpha are in GP, the only possible values are when both sin⁡α\sin \alpha and cos⁡2α\cos 2 \alpha are 1 or -1, leading to α=nπ+(−1)n−1π2\alpha = n \pi+(-1)^{n-1} \frac{\pi}{2}.
117 Medium
117 of 294
Application +4 | -1
If 2sin⁡2(x−π3)−5sin⁡(x−π3)+2<02 \sin ^2\left(x-\frac{\pi}{3}\right)-5 \sin \left(x-\frac{\pi}{3}\right)+2 < 0, then xx belongs to-
View answer and explanation
Correct option: D
((4n+1)π2,(12n+7)π6),n∈I\left((4 n+1) \frac{\pi}{2},(12 n+7) \frac{\pi}{6}\right), n \in I
Option D (((4n+1)π2,(12n+7)π6),n∈I((4 n+1) \frac{\pi}{2},(12 n+7) \frac{\pi}{6}), n \in I) is correct because it represents the solution set for x after solving the given trigonometric inequality.
118 Medium
118 of 294
Analyzing +4 | -1
Let A={θ:sin⁡(θ)=tan⁡(θ)}{A}=\{\theta: \sin (\theta)=\tan (\theta)\} and B={θ:cos⁡(θ)=1}{B}=\{\theta: \cos (\theta)=1\} be two sets. Then :
View answer and explanation
Correct option: B
A⊄BA \not \subset B
The correct answer is A ⊄B\not\subset B because set A contains elements that are not in set B, specifically odd multiples of \π\pi.
119 Medium
119 of 294
Analyzing +4 | -1
If sin⁡θ=3sin⁡(θ+2α)\sin \theta=3 \sin (\theta+2 \alpha), then the value of tan⁡(θ+α)+2tan⁡α\tan (\theta+\alpha)+2 \tan \alpha is
View answer and explanation
Correct option: D
0
Option D (0) is correct because applying componendo and dividendo and simplifying leads to the given expression equaling zero.
120 Medium
120 of 294
Application +4 | -1
If sin⁡2x=4cos⁡x\sin 2 x=4 \cos x, then xx is equal to
View answer and explanation
Correct option: D
2nπ±π2,n∈Z2 n \pi \pm \frac{\pi}{2}, n \in Z
Option D (2nπ±π2,n∈Z2 n \pi \pm \frac{\pi}{2}, n \in Z) is correct because simplifying the equation sin⁡2x=4cos⁡x\sin 2x = 4 \cos x, we get cos⁡x=0\cos x = 0, which leads to the solution x=2nπ±π2x = 2 n \pi \pm \frac{\pi}{2}.
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