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JEE • MATHEMATICS • TRIGONOMETRIC EQUATIONS

Trigonometric equations practice · Mixed difficulty

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211 Easy
211 of 581
Application +4 | -1
If sin⁡2  θ=14{\sin ^2}\,\,\mathrm{\theta } = \frac{1}{4} , then the most general value of θ\mathrm{\theta } is
View answer and explanation
Correct option: C
nπ  ±  π6n\mathrm{\pi }\,\, \pm \,\,\frac{\mathrm{\pi }}{6}
Option C (nπ;±;π6n\mathrm{\pi }\\; \pm \\; {\rm{\frac{{\pi }}{6}}}) is correct because it represents the general solution for the given trigonometric equation.
212 Medium
212 of 581
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}.
213 Medium
213 of 581
Application +4 | -1
The value of 2cot⁡2(π/6)+4tan⁡2(π/6)−3cosec⁡(π/6)2 \cot ^2(\pi / 6)+4 \tan ^2(\pi / 6)-3 \operatorname{cosec}(\pi / 6) is
View answer and explanation
Correct option: C
04-Mar
The correct answer is 4/3. This is obtained by substituting the values of cot, tan, and cosec at π/6\pi/6 and simplifying the expression.
214 Easy
214 of 581
Analyzing +4 | -1
If the equation cos⁡  3x  cos⁡3  x+sin⁡  3x  sin⁡3  x=0\cos \,\,3x\,\,{\cos ^3}\,\,x + \sin \,\,3x\,\,{\sin ^3}\,\,x = 0, then xx is equal to
View answer and explanation
Correct option: A
(2n+1)  π4\left( {2n + 1} \right)\,\,\frac{\pi }{4}
Option A ((2n+1)π4\left( {2n + 1} \right)\frac{\pi }{4}) is correct because it is derived from simplifying the trigonometric equation and solving for x.
215 Medium
215 of 581
Application +4 | -1
The number of roots of the equation 3  sin⁡2x=8  cos⁡  x3\,\,{\sin ^2}x = 8\,\,\cos \,\,x in (  −π2,  π2)\left( {\; - \frac{\mathrm{\pi }}{2},\,\,\frac{\mathrm{\pi }}{2}} \right) is
View answer and explanation
Correct option: A
1\mathrm{1}
Option A (1\mathrm{1}) is correct because by solving the equation 3 sin⁡2x=8 cos⁡ x3\,{\sin ^2}x = 8\,\cos \,x in the interval (  −π2, π2)\left( {\; - \frac{\mathrm{\pi }}{2},\,\frac{\mathrm{\pi }}{2}} \right), we find that there is only one root.
216 Medium
216 of 581
Analyzing +4 | -1
If (asin⁡4θ+b)(a+b)=abcos⁡4θ\left(a \sin ^4 \theta+b\right)(a+b)=a b \cos ^4 \theta, then
View answer and explanation
Correct option: A
∣a∣≥∣b∣|a| \geq|b|
Option A (∣a∣≥∣b∣|a| \geq |b|) is correct because it is derived from the given equation and trigonometric identities.
217 Medium
217 of 581
Analyzing +4 | -1
The number of solutions of cos⁡x=∣1+sin⁡x∣\cos x=|1+\sin x|, 0≤x≤3π0 \leq x \leq 3 \pi is
View answer and explanation
Correct option: A
3
Option A (3) is correct because the equation cos(x) = |1 + sin(x)| has three solutions within the interval 0 <= x <= 3*pi.
218 Medium
218 of 581
Analyzing +4 | -1
If sin⁡2x+asin⁡x+1=0\sin ^2 x+a \sin x+1=0 has no real number solution, then
View answer and explanation
Correct option: C
∣a∣<2|a| < 2
Option C (∣a∣<2|a| < 2) is correct because it is derived from the condition for no real solution of the given equation.
219 Medium
219 of 581
Application +4 | -1
The maximum value of 27cos⁡2x81sin⁡2x27^{\cos 2 x} 81^{\sin 2 x} is
View answer and explanation
Correct option: B
353^5
Option B (353^5) is correct because the maximum value is obtained when 3cos⁡2x+4sin⁡2x3 \cos 2x + 4 \sin 2x is maximized, which occurs at 32+42=5\sqrt{3^2 + 4^2} = 5, resulting in 353^5.
220 Medium
220 of 581
Analyzing +4 | -1
If sin⁡2(θ−α)cos⁡α=cos⁡2(θ−α)sin⁡α=msin⁡αcos⁡α\sin ^2(\theta-\alpha) \cos \alpha=\cos ^2(\theta-\alpha) \sin \alpha =m \sin \alpha \cos \alpha, then
View answer and explanation
Correct option: B
∣m∣≥12|m| \geq \frac{1}{\sqrt{2}}
Option B (∣m∣≥12|m| \geq \frac{1}{\sqrt{2}}) is correct because it is derived by manipulating the given trigonometric equations and using the range of the sine function.
221 Medium
221 of 581
Application +4 | -1
If cos⁡x+cos⁡y+cos⁡α=0\cos x+\cos y+\cos \alpha=0 and sin⁡x+sin⁡y+sin⁡α=0\sin x+\sin y+\sin \alpha=0, then cot⁡(x+y2)\cot \left(\frac{x+y}{2}\right) is equal to
View answer and explanation
Correct option: C
cot⁡α\cot \alpha
Option C (cot⁡α\cot \alpha) is correct because it is derived by dividing the sum-to-product forms of the given equations and simplifying.
222 Medium
222 of 581
Application +4 | -1
If tan⁡  θ+tan⁡  4  θ+tan⁡7  θ=tan⁡  θtan⁡  4  θ  tan⁡7  θ\tan \,\,\theta + \tan \,\,4\;\theta + \tan 7\;\theta = \tan \,\,\theta \tan \,\,4\;\theta \,\,\tan 7\;\theta, then θ=\theta =
View answer and explanation
Correct option: C
n  π12,  n∈Z\frac{{n\;\pi }}{{12}},\,\,n \in Z
Option C (n;π12,,n∈Z\frac{{n\\;\pi }}{{12}},\\,n \in Z) is correct because it is the solution to the trigonometric equation obtained by using the tangent addition formula.
223 Easy
223 of 581
Application +4 | -1
If θ1\theta_1 and θ2\theta_2 are two values lying in [0,2π][0,2 \pi] for which tan⁡θ=λ\tan \theta=\lambda, then tan⁡θ12tan⁡θ22\tan \frac{\theta_1}{2} \tan \frac{\theta_2}{2} is equal to
View answer and explanation
Correct option: B
−1-1
Option B (-1) is correct because the product of the roots of the quadratic equation in tan⁡θ2\tan \frac{{\theta}}{{2}} is -1.
224 Hard
224 of 581
Analyzing +4 | -1
Equation cos⁡  2x+7=a(2−sin⁡ x)\cos \,\,2x + 7 = a\left( {2 - \sin \,x} \right) can have a real solution for
View answer and explanation
Correct option: B
a∈[2,   6]a \in \left[ {2,\;\,6} \right]
Option B (a∈[2,6]a \in [2, 6]) is correct because the equation has a real solution only when the value of 'a' falls within this interval, ensuring that sin⁡x\sin x remains within its valid range [-1, 1].
225 Easy
225 of 581
Analyzing +4 | -1
If sin⁡ 3  θ=4 sin⁡θ   (sin⁡2x−sin⁡2θ),  θ≠n  π,  n∈Z\sin \,3\;\theta = 4\,\sin \theta \;\,\left( {{{\sin }^2}x - {{\sin }^2}\theta } \right),\,\,\theta \ne n\;\pi ,\,\,n \in Z. Then, the set of values of xx is
View answer and explanation
Correct option: A
{n  π  ±  π3  :n∈Z}\left\{ {n\;\pi \,\, \pm \,\,\frac{\pi }{3}\;:n \in Z} \right\}
Option A ({n;π,±,π3;:n∈Z}\{ {n\\;\pi \\,\pm \\,\frac{\pi }{{3}}}\\;:n \in Z \}) is correct because it is derived by simplifying the given trigonometric equation and solving for x.

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226 Medium
226 of 581
Application +4 | -1
If α+β−γ=π\alpha+\beta-\gamma=\pi, then sin⁡2α+sin⁡2β−sin⁡2γ\sin ^2 \alpha+\sin ^2 \beta-\sin ^2 \gamma is equal to
View answer and explanation
Correct option: A
2sin⁡αsin⁡βcos⁡γ2 \sin \alpha \sin \beta \cos \gamma
Option A (2sin⁡αsin⁡βcos⁡γ2 \sin \alpha \sin \beta \cos \gamma) is correct because it is the simplified form of the given expression using trigonometric identities and the given condition α+β−γ=π\alpha+\beta-\gamma=\pi.
227 Medium
227 of 581
Analyzing +4 | -1
If tan⁡x=ntan⁡y,n∈R+\tan x=n \tan y, n \in R^{+}, then maximum value of sec⁡2(x−y)\sec ^2(x-y) is
View answer and explanation
Correct option: D
(n+1)24n\frac{(n+1)^2}{4 n}
Option D ((n+1)24n\frac{(n+1)^2}{4 n}) is correct because the maximum value of sec⁡2(x−y)\sec^2(x-y) is found by using trigonometric identities and maximizing the expression with respect to tan⁡y\tan y.
228 Hard
228 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
View answer and explanation
Correct option: B
−(4+7)3- \frac{{\left( {4 + \sqrt 7 } \right)}}{3}
Option B (−(4+7)3- \frac{{\left( {4 + \sqrt 7 } \right)}}{3}) is correct because solving the given equation for tan⁡x\tan x within the specified interval yields this value.
229 Medium
229 of 581
Analyzing +4 | -1
The equation sin⁡4x−2cos⁡2x+a2=0{\sin ^4}x - 2{\cos ^2}x + {a^2} = 0 is solvable for
View answer and explanation
Correct option: B
− 2≤a≤2- \,\sqrt 2 \le a \le \sqrt 2
Option B (−2≤a≤2- \sqrt 2 \le a \le \sqrt 2) is correct because it defines the range of 'a' for which the given equation has real solutions.
230 Medium
230 of 581
Application +4 | -1
If sin⁡x+sin⁡2x=1\sin x+\sin ^2 x=1, then cos⁡8x+2cos⁡6x+cos⁡4x\cos ^8 x+2 \cos ^6 x+\cos ^4 x is equal to
View answer and explanation
Correct option: C
1
Option C (1) is correct because by using the given equation and trigonometric identities, the expression simplifies to 1.
231 Easy
231 of 581
Application +4 | -1
The equation 3  sin⁡2x+10  cos⁡  x−6=03\,\,{\sin ^2}x + 10\,\,\cos \,\,x - 6 = 0 is satisfied, if
View answer and explanation
Correct option: B
x=2nπ  ±  cos⁡−1(13)x = 2n\pi \,\, \pm \,\,{\cos ^{ - 1}}\left( {\frac{1}{3}} \right)
Option B (x=2nπ ± cos⁡−1(13)x = 2n\pi \, \pm \, {\cos ^{ - 1}}(\frac{1}{3})) is correct because it represents the general solution to the trigonometric equation 3sin⁡2x+10cos⁡x−6=03 \sin^2 x + 10 \cos x - 6 = 0 after simplification and solving for x.
232 Easy
232 of 581
Analyzing +4 | -1
The expression cos⁡2ϕ+cos⁡2(a+ϕ)−2cos⁡acos⁡ϕcos⁡(a+ϕ)\cos ^2 \phi+\cos ^2(a+\phi)-2 \cos a \cos \phi \cos (a+\phi) is independent of
View answer and explanation
Correct option: A
ϕ\phi
Option A (ϕ\phi) is correct because the expression simplifies to sin⁡2a\sin^2 a, which is independent of ϕ\phi.
233 Medium
233 of 581
Analyzing +4 | -1
If sin⁡θ=x2+y2x2−y2\sin \theta=\frac{x^2+y^2}{x^2-y^2}, then
View answer and explanation
Correct option: C
x=0x=0 or y=0y=0
Option C (x=0x=0 or y=0y=0) is correct because the given condition sin⁡θ=x2+y2x2−y2\sin \theta=\frac{{x^2+y^2}}{{x^2-y^2}} implies ∣x2+y2x2−y2∣≤1\left|\frac{{x^2+y^2}}{{x^2-y^2}}\right| \leq 1, which leads to x=0x=0 or y=0y=0.
234 Medium
234 of 581
Application +4 | -1
The most general values of θ\theta satisfying tan⁡  θ+tan⁡  (3 π4+θ)=2\tan \,\,\theta + \tan \,\,\left( {\frac{{3\,\pi }}{4} + \theta } \right) = 2 are given by
View answer and explanation
Correct option: B
nπ+π3,   n∈Zn\pi + \frac{\pi }{3},\;\,n \in Z
Option B (nπ+π3,n∈Zn\pi + \frac{\pi}{3}, n \in Z) is correct because it represents the general solution of the given trigonometric equation.
235 Easy
235 of 581
Analyzing +4 | -1
If x≠nπ2x \neq \frac{n \pi}{2} and (cos⁡x)sin⁡2x−3sin⁡x+2=1(\cos x)^{\sin ^2 x-3 \sin x+2}=1, then all solutions of xx is/are given by
View answer and explanation
Correct option: D
None of these
Option A (6) is correct because there are 6 possible ordered pairs (x, y) that satisfy the given equation within the specified intervals.
236 Medium
236 of 581
Analyzing +4 | -1
If x∈(π,2π)x \in(\pi, 2 \pi) and 1+cos⁡x+1−cos⁡x1+cos⁡x−1−cos⁡x=cot⁡(a+x2)\frac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}}=\cot \left(a+\frac{x}{2}\right), then aa is equal to
View answer and explanation
Correct option: A
π4\frac{\pi}{4}
Option A (π4\frac{\pi}{4}) is correct because simplifying the given expression leads to cot⁡(π4+x2)\cot(\frac{\pi}{4} + \frac{x}{2}).
237 Easy
237 of 581
Application +4 | -1
The solution of equation cos⁡2θ+sin⁡θ+1=0{\cos ^2}\theta + \sin \theta + 1 = 0 lies in the interval
View answer and explanation
Correct option: D
(5π4,   7π4)\left( {\frac{{5\pi }}{4},\;\,\frac{{7\pi }}{4}} \right)
Option D ((5π4,  7π4)\left( {\frac{{5\pi }}{4},\;\frac{{7\pi }}{4}} \right)) is correct because the solution to the equation cos⁡2θ+sin⁡θ+1=0{\cos ^2}\theta + \sin \theta + 1 = 0 is θ=3π2\theta = \frac{{3\pi }}{2}, which lies in this interval.
238 Easy
238 of 581
Application +4 | -1
cos⁡(π4−x)cos⁡(π4−y)−sin⁡(π4−x)sin⁡(π4−y)\cos \left( {\frac{\pi }{4} - x} \right)\cos \left( {\frac{\pi }{4} - y} \right) - \sin \left( {\frac{\pi }{4} - x} \right)\sin \left( {\frac{\pi }{4} - y} \right) is equal to
View answer and explanation
Correct option: B
sin⁡(x+y)\sin \left( {x + y} \right)
Option C (cos⁡(x+y)\cos \left( {x + y} \right)) is correct because it is the simplified form of the given expression using the cosine addition formula.
239 Easy
239 of 581
Application +4 | -1
If tan⁡  (απ4)=cot⁡  (βπ4)\tan \,\,\left( {\frac{{\alpha \pi }}{4}} \right) = \cot \,\,\left( {\frac{{\beta \pi }}{4}} \right), then
View answer and explanation
Correct option: D
α+β=2(2n+1),  n∈Z\alpha + \beta = 2\left( {2n + 1} \right),\;n \in Z
Option D (α+β=2(2n+1),  n∈Z\alpha + \beta = 2\left( {2n + 1} \right),\;n \in Z) is correct because by converting cotangent to tangent and using the general solution for tangent equations, we find the relationship between α\alpha and β\beta.
240 Easy
240 of 581
Application +4 | -1
The most general value of θ\theta which satisfy both the equations cos⁡  θ=−12\cos \,\,\theta = - \frac{1}{{\sqrt 2 }} and tan⁡  θ=1\tan \,\,\theta = 1, is
View answer and explanation
Correct option: A
2nπ+5π4,    n∈I2n\pi + \frac{{5\pi }}{4},\;\,\,n \in I
Option A is correct because it identifies the general solution that satisfies both the cosine and tangent conditions, considering the quadrant in which both conditions are met.
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