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JEE • MATHEMATICS • TRIGONOMETRIC EQUATIONS

Trigonometric equations practice · Easy

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151 Easy
151 of 261
Analyzing +4 | -1
The maximum value of sin⁡(cos⁡x)\sin (\cos x) is
View answer and explanation
Correct option: A
sin⁡1\sin 1
Option A (sin⁡1\sin 1) is correct because the maximum value of cos⁡x\cos x is 1, and sin⁡x\sin x is increasing on [−1,1][-1, 1].
152 Easy
152 of 261
Application +4 | -1
If 3sin⁡θ+4cos⁡θ=53 \sin \theta+4 \cos \theta=5, then the value of 3cos⁡θ−4sin⁡θ3 \cos \theta-4 \sin \theta is equal to
View answer and explanation
Correct option: A
00
Option A (0) is correct because by squaring both sides of the given equation and using trigonometric identities, we find that (3cos⁡θ−4sin⁡θ)2=0(3 \cos \theta - 4 \sin \theta)^2 = 0.
153 Easy
153 of 261
Application +4 | -1
The root of the equation 1−cos⁡  θ=sin⁡  θ .  sin⁡  θ21 - \cos \,\,\theta = \sin \,\,\theta \,.\,\,\sin \,\,\frac{\theta }{2} is
View answer and explanation
Correct option: B
2k  π,   k∈I2k\;\pi ,\;\,k \in I
Option B (2k  π,   k∈I2k\;\pi ,\;\,k \in I) is correct because it represents the general solution of the given trigonometric equation.
154 Easy
154 of 261
Understanding +4 | -1
In △ABC\triangle A B C, the value of the expression cosec⁡A(sin⁡Bcos⁡C+cos⁡Bsin⁡C)\operatorname{cosec} A(\sin B \cos C+\cos B \sin C) is
View answer and explanation
Correct option: A
1
Option A (1) is correct because the expression simplifies to cosec⁡Asin⁡(B+C)=cosec⁡Asin⁡(π−A)=cosec⁡Asin⁡A=1\operatorname{cosec} A \sin(B+C) = \operatorname{cosec} A \sin(\pi - A) = \operatorname{cosec} A \sin A = 1.
155 Easy
155 of 261
Application +4 | -1
The number of values of xx in [0,   2π]\left[ {0,\;\,2\pi } \right] satisfying the equation 3  cos⁡  2x−10  cos⁡  x+7=03\,\,\cos \,\,2x - 10\,\,\cos \,\,x + 7 = 0 is

View answer and explanation
Correct option: D
4\mathrm{4}
Option D (4) is correct because solving the trigonometric equation yields four solutions for x in the given interval [0, 2π].
156 Easy
156 of 261
Analyzing +4 | -1
If 2cos⁡2x2sin⁡2x=x2+x−22 \cos ^2 \frac{x}{2} \sin ^2 x=x^2+x^{-2},0<x≤π20 < x \leq \frac{\pi}{2}, then xx is equal to
View answer and explanation
Correct option: A
ϕ\phi
Option A (ϕ\phi) is correct because the LHS is always less than 2, while the RHS is always greater than or equal to 2, so there is no solution.
157 Easy
157 of 261
Analyzing +4 | -1
If θ∈  [0,  5π]\theta \in \,\,\left[ {0,\,\,5\pi } \right] and r∈Rr \in R such that 2  sin⁡  θ=r4−2r2+32\,\,\sin \,\,\theta = {r^4} - 2{r^2} + 3 then the maximum number of values of the pair (r,  θ)\left( {r,\,\,\theta } \right) is
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Correct option: A
6\mathrm{6}
Option A (6) is correct because the equation 2 sin θ = r4 - 2r2 + 3 has solutions only when both sides equal 2, leading to 6 possible pairs of (r, θ).
158 Easy
158 of 261
Analyzing +4 | -1
If sin⁡2θ−2sin⁡θ−1=0\sin ^2 \theta-2 \sin \theta-1=0 is to be satisfied for exactly 4 distinct values of θ∈[0,nπ],n∈N\theta \in[0, n \pi], n \in N, then the least value of nn is
View answer and explanation
Correct option: C
4
Option C (4) is correct because the equation sin⁡2θ−2sin⁡θ−1=0\sin^2 \theta - 2 \sin \theta - 1 = 0 has exactly 4 distinct values of θ\theta in the interval [0,nπ][0, n\pi] when n = 4.
159 Easy
159 of 261
Application +4 | -1
The values of θ\theta satisfying sin⁡  7  θ=sin⁡  4  θ−sin⁡  θ\sin \,\,7\;\theta = \sin \,\,4\;\theta - \sin \,\,\theta and 0<θ<π20 < \theta < \frac{\pi }{2} are
View answer and explanation
Correct option: A
π9,  π4\frac{\pi }{9},\,\,\frac{\pi }{4}
Option A (π9,  π4\frac{\pi }{9},\,\,\frac{\pi }{4}) is correct because these are the solutions to the trigonometric equation within the given range, found by using trigonometric identities and solving for θ\theta.
160 Easy
160 of 261
Application +4 | -1
The values of θ\theta in the interval (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) satisfying the equation (3)sec⁡2θ=tan⁡4θ+2tan⁡2θ(\sqrt{3})^{\sec ^2 \theta}=\tan ^4 \theta+2 \tan ^2 \theta is
View answer and explanation
Correct option: A
π4\frac{\pi}{4}
Option A (π4\frac{\pi}{4}) is correct because it is the only value in the given interval that satisfies the provided trigonometric equation.
161 Easy
161 of 261
Analyzing +4 | -1
Total number of solutions of cos⁡  x=1−sin⁡2x\cos \,\,x = \sqrt {1 - \sin 2x} in [0,   2π]\left[ {0,\;\,2\mathrm{\pi }} \right] is equal to
View answer and explanation
Correct option: A
2
Option A (2) is correct because there are two solutions to the given equation within the specified interval.
162 Easy
162 of 261
Analyzing +4 | -1
The arithmetic mean of the roots of the equation 4cos⁡3x−4cos⁡2x−cos⁡(315 π+x)=14{\cos ^3}x - 4{\cos ^2}x - \cos \left( {315\,\pi + x} \right) = 1 in the interval (0,315)\left( {\mathrm{0, 315}} \right) is equal to
View answer and explanation
Correct option: B
51 π\mathrm{51}\,\mathrm{\pi }
Option B (51 π\mathrm{51}\,\mathrm{\pi }) is correct because it is the arithmetic mean of the roots of the given equation within the specified interval.
163 Easy
163 of 261
Application +4 | -1
The exhaustive range of values of θ\theta in the interval (0,π)(0, \pi) such that the points (3,5)(3,5) and (3sin⁡θ,cos⁡θ)(\sqrt{3} \sin \theta, \cos \theta) lies on the same side of line x+y−1=0x+y-1=0 is
View answer and explanation
Correct option: A
3
The correct answer is that the range of θ is (0, 2π/3). This is found by ensuring that both points lie on the same side of the line x+y-1=0, leading to the inequality sin⁡(θ+π/6)>1/2\sin(\theta + \pi/6) > 1/2.
164 Easy
164 of 261
Analyzing +4 | -1
If   tan⁡  x+cot⁡  x=2\;\tan \,\,x + \cot \,\,x = 2 then sin⁡2n  x+cos⁡2n  x  {\sin ^{2n}}\,\,x + {\cos ^{2n}}\,\,x\; is equal to
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the expression simplifies to 12n−1\frac{1}{2^{n-1}}, which is not among the given options.
165 Easy
165 of 261
Application +4 | -1
The value of 6(sin⁡6θ+cos⁡6θ)−9(sin⁡4θ+cos⁡4θ)+46\left(\sin ^6 \theta+\cos ^6 \theta\right)-9\left(\sin ^4 \theta+\cos ^4 \theta\right)+4 is
View answer and explanation
Correct option: C
1
The correct answer is 1 because the expression simplifies to 1 using trigonometric identities and algebraic manipulation.

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166 Easy
166 of 261
Application +4 | -1
If sin⁡θ+cos⁡θ=0\sin \theta+\cos \theta=0 and 0<θ<π0 < \theta < \pi, then θ\theta is equal to
View answer and explanation
Correct option: D
3π4\frac{3 \pi}{4}
Option D (3π4\frac{3 \pi}{4}) is correct because it satisfies the equation sin⁡θ+cos⁡θ=0\sin \theta + \cos \theta = 0 within the given range 0<θ<π0 < \theta < \pi.
167 Easy
167 of 261
Application +4 | -1
If tan⁡2  θ−(1+3)tan⁡  θ+3=0{\tan ^2}\,\,\mathrm{\theta } - \left( {1 + \sqrt 3 } \right)\tan \,\,\mathrm{\theta } + \sqrt 3 = 0, then the general value of θ\mathrm{\theta } is
View answer and explanation
Correct option: A
nπ  +  π4,   nπ  +  π3n\pi \,\, + \,\,\frac{\pi }{4},\;\,n\pi \,\, + \,\,\frac{\pi }{3}
Option A (nπ;+;π4,;nπ;+;π3n\pi \\; + \\; \frac{\pi }{{4}},\\;n\pi \\; + \\; \frac{\pi }{{3}}) is correct because it represents the general solutions obtained by solving the quadratic equation in tan⁡θ\tan \theta.
168 Easy
168 of 261
Application +4 | -1
The general value of θ\theta satisfying the equation 2  sin⁡2  θ−3  sin⁡  θ−2=02\,\,{\sin ^2}\,\,\theta - 3\,\,\sin \,\,\theta - 2 = 0 is
View answer and explanation
Correct option: A
nπ+(−1)n+1  π6n\pi + {\left( { - 1} \right)^{n + 1}}\,\,\frac{\pi }{6}
Option A (nπ+(−1)n+1 π6n\pi + {\left( { - 1} \right)^{n + 1}}\,\frac{\pi }{6}) is correct because it represents the general solution to the trigonometric equation 2sin⁡2 θ−3 sin⁡ θ−2=02{\sin ^2}\,\theta - 3\,\sin \,\theta - 2 = 0.
169 Easy
169 of 261
Application +4 | -1
The number of solutions of cos⁡  2θ=sin⁡θ\cos \,\,2\theta = \sin \theta in (0,   2π)\left( {0,\,\;2\pi } \right) is
View answer and explanation
Correct option: C
3
Option C (3) is correct because there are three solutions to the equation cos⁡2θ=sin⁡θ\cos 2\theta = \sin \theta in the interval (0,2π)(0, 2\pi).
170 Easy
170 of 261
Application +4 | -1
The general solution of the equation 2sin⁡2x+3cos⁡x+1=02 \sin ^2 x+\sqrt{3} \cos x+1=0 is
View answer and explanation
Correct option: C
x=2nπ±5π6x=2 n \pi \pm \frac{5 \pi}{6}
Option C (x=2nπ±5π6x=2 n \pi \pm \frac{{5 \pi}}{{6}}) is the correct general solution obtained by solving the trigonometric equation for cos(x).
171 Easy
171 of 261
Analyzing +4 | -1
If sin⁡θ=x+ax,∀ x∈R∼{0}\sin \theta=x+\frac{a}{x}, \forall \, x \in R \sim\{0\}, then
View answer and explanation
Correct option: C
a≤14a \leq \frac{1}{4}
Option C (a≤14a \leq \frac{1}{4}) is correct because it is derived from the condition for the existence of real roots of the quadratic equation.
172 Easy
172 of 261
Application +4 | -1
The solution of the inequality −8≤5x−3<7-8 \leq 5 x-3 < 7 is [−a,b).[-a, b). Sum of &#39;aa&#39; and &#39;bb&#39; is
View answer and explanation
Correct option: A
3
Option A (3) is correct because solving the inequality −8≤5x−3<7-8 \leq 5x - 3 < 7 results in −1≤x<2-1 \leq x < 2, and the sum of the absolute values of the endpoints is 3.
173 Easy
173 of 261
Analyzing +4 | -1
If A+B=π3A+B=\frac{\pi}{3}, where A,B∈R+A, B \in R^{+}, then the minimum value of sec⁡A+sec⁡B\sec A+\sec B is
View answer and explanation
Correct option: B
43\frac{4}{\sqrt{3}}
Option B (43\frac{4}{\sqrt{3}}) is correct because the minimum value of sec⁡A+sec⁡B\sec A + \sec B is found using the property that the average of secants is greater than or equal to the secant of the average angle.
174 Easy
174 of 261
Application +4 | -1
{x∈R  :cos⁡  2x+2  cos⁡2x=2}\left\{ {x \in R\;:\cos \,\,2x + 2\,\,{{\cos }^2}x = 2} \right\} is equal to

View answer and explanation
Correct option: B
{nπ  ±  π6  :n∈Z}\left\{ {n\pi \,\, \pm \,\,\frac{\pi }{6}\;:n \in Z} \right\}
Option D (3/4) is correct because by using trigonometric identities and simplifying the expression, we can find the value to be 3/4.
175 Easy
175 of 261
Analyzing +4 | -1
The number of solutions of the equation tan⁡  θ+sec⁡  θ=2  cos⁡  θ\tan \,\,\theta + \sec \,\,\theta = 2\,\,\cos \,\,\theta lying in the interval [0,   2  π]\left[ {0,\;\,2\;\pi } \right], is

View answer and explanation
Correct option: C
2\mathrm{2}
Option C (2) is correct because solving the equation tan⁡θ+sec⁡θ=2cos⁡θ\tan \theta + \sec \theta = 2 \cos \theta yields two valid solutions within the interval [0,2π][0, 2\pi].
176 Easy
176 of 261
Application +4 | -1
If 2cos⁡θ−sin⁡θ+2=02 \cos \theta-\sin \theta+2=0, then the value of 10cos⁡θ+15sin⁡θ10 \cos \theta+15 \sin \theta is equal to:
View answer and explanation
Correct option: B
6
Option B (6) is correct because by solving the given trigonometric equation and substituting the values of cos⁡θ\cos \theta and sin⁡θ\sin \theta into the expression 10cos⁡θ+15sin⁡θ10 \cos \theta + 15 \sin \theta, we get 6.
177 Easy
177 of 261
Application +4 | -1
The solution set of the inequality 4x+3<6x+74 x+3 < 6 x+7 is (−a,∞)(-a, \infty). The value of &#39;aa&#39; is
View answer and explanation
Correct option: A
2
Option A (2) is correct because solving the inequality 4x+3<6x+74x + 3 < 6x + 7 results in x>−2x > -2, which corresponds to the interval (−2,∞)(-2, \infty).
178 Easy
178 of 261
Analyzing +4 | -1
If equation sin⁡4x=1+tan⁡8x\sin ^4 x=1+\tan ^8 x, then xx is equal to
View answer and explanation
Correct option: A
ϕ\phi
Option A (ϕ\phi) is correct because the equation sin⁡4x=1+tan⁡8x\sin^4 x = 1 + \tan^8 x has no solution since sin⁡4x\sin^4 x cannot be 1 and tan⁡8x\tan^8 x be 0 simultaneously.
179 Easy
179 of 261
Analyzing +4 | -1
The number of solutions of the equation 1+sin⁡x  sin⁡2x2=01 + \sin x\,\,{\sin ^2}\frac{x}{2} = 0 in [− π, π]\left[ { - \,\mathrm{\pi },\,\mathrm{\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 simplifies to sin⁡2x−2sin⁡x=4\sin 2x - 2 \sin x = 4, which has no solutions since the range of sin⁡\sin is [-1, 1].
180 Easy
180 of 261
Application +4 | -1
If tan⁡(k+1)θ=tan⁡  θ\tan \left( {k + 1} \right)\theta = \tan \,\,\theta, then θ\theta belongs to the set
View answer and explanation
Correct option: C
{nπ  /  k  :n∈I}\left\{ {n\pi \;/\;k\;:n \in I} \right\}
Option C ({nπ  /  k  :  n∈I}\{n\pi \;/\;k\;:\;n \in I\}) is correct because it represents the general solution for θ\theta derived from the given trigonometric equation.
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