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JEE • MATHEMATICS • TRIGONOMETRIC EQUATIONS

Trigonometric equations practice · Medium

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121 Medium
121 of 294
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.
122 Medium
122 of 294
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.
123 Medium
123 of 294
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.
124 Medium
124 of 294
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.
125 Medium
125 of 294
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.
126 Medium
126 of 294
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.
127 Medium
127 of 294
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.
128 Medium
128 of 294
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.
129 Medium
129 of 294
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.
130 Medium
130 of 294
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.
131 Medium
131 of 294
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.
132 Medium
132 of 294
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.
133 Medium
133 of 294
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.
134 Medium
134 of 294
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.
135 Medium
135 of 294
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.

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136 Medium
136 of 294
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}).
137 Medium
137 of 294
Analyzing +4 | -1
The solution of trigonometric equation cos⁡4x+sin⁡4x=2cos⁡(2x+π)cos⁡(2x−π)\cos ^4 x+\sin ^4 x=2 \cos (2 x+\pi) \cos (2 x-\pi) is
View answer and explanation
Correct option: B
x=nπ4+(−1)n4sin⁡−1(±223)x=\frac{n \pi}{4}+\frac{(-1)^n}{4} \sin ^{-1}\left(\frac{ \pm 2 \sqrt{2}}{3}\right)
Option B (x=nπ4+(−1)n4sin⁡−1(±223)x=\frac{{n \pi}}{{4}}+\frac{{(-1)^n}}{{4}} \sin^{-1}(\frac{{ \pm 2 \sqrt{{2}}}}{{3}})) is correct because simplifying the given equation leads to this general solution.
138 Medium
138 of 294
Analyzing +4 | -1
If cos⁡22x−sin⁡4x−2cos⁡2x=λ\cos ^2 2 x-\sin ^4 x-2 \cos ^2 x=\lambda, has a solution ∀ x∈R\forall \, x \in R, then the range of λ\lambda is
View answer and explanation
Correct option: B
[−43,0]\left[-\frac{4}{3}, 0\right]
The correct answer is B ([−43,0]\left[-\frac{4}{3}, 0\right]) because it represents the range of the function after simplification and completing the square.
139 Medium
139 of 294
Understanding +4 | -1
Assertion: The inequality ax+by<0a x+b y < 0 is strict inequality.

Reason: The inequality ax+b≥0a x+b \geq 0 is slack inequality.
View answer and explanation
Correct option: B
Assertion is correct, reason is correct; reason is not a
correct explanation for assertion
Option B (Assertion is correct, reason is correct; reason is not a correct explanation for assertion) is correct because both statements are true, but the reason doesn't explain the assertion.
140 Medium
140 of 294
Application +4 | -1
If in any △ABC\triangle A B C, the value of ∠A\angle A is obtained from the equation 3cos⁡A+2=03 \cos A+2=0, then the equation whose roots are sin⁡A\sin A and tan⁡A\tan A, is
View answer and explanation
Correct option: A
6x2+5x−5=06 x^2+\sqrt{5} x-5=0
Option A (6x2+5x−5=06 x^2+\sqrt{5} x-5=0) is correct because it represents the quadratic equation whose roots are sin⁡A\sin A and tan⁡A\tan A, derived from the given trigonometric relationship.
141 Medium
141 of 294
Analyzing +4 | -1
The general solution of sin⁡3x+sin⁡x−3sin⁡2x\sin 3 x+\sin x-3 \sin 2 x =cos⁡3x+cos⁡x−3cos⁡2x=\cos 3 x+\cos x-3 \cos 2 x is
View answer and explanation
Correct option: A
nπ2+π8\frac{n \pi}{2}+\frac{\pi}{8} for nn integer
Option A is correct because the general solution is found by simplifying the trigonometric equation and solving for x.
142 Medium
142 of 294
Analyzing +4 | -1
The number of x∈[0,2π]x \in[0,2 \pi] for which
∣2sin⁡4x+18cos⁡2x−2cos⁡4x+18sin⁡2x∣=1 is \left|\sqrt{2 \sin ^{4} x+18 \cos ^{2} x}-\sqrt{2 \cos ^{4} x+18 \sin ^{2} x}\right|=1 \text { is }
View answer and explanation
Correct option: D
8
Option D (8) is correct because squaring both sides and solving the equation gives 8 solutions in the interval [0, 2π\pi].
143 Medium
143 of 294
Analyzing +4 | -1
If 2tan⁡2x−5sec⁡x2 \tan ^2 x-5 \sec x is equal to 1 for exactly 7 distinct values of x∈[0,nπ2]n∈Nx \in\left[0, \frac{n \pi}{2}\right] n \in N, then the greatest value of nn is
View answer and explanation
Correct option: D
15
Option D (15) is correct because the equation simplifies to sec(x) = 3, which yields 7 distinct solutions within the given interval, resulting in n=15.
144 Medium
144 of 294
Application +4 | -1
Find the value of cos⁡(29π/3)\cos (29 \pi / 3).
View answer and explanation
Correct option: D
1/21 / 2
Option D (1/2) is correct because by simplifying the angle and using trigonometric identities, cos⁡(29π/3)\cos (29 \pi / 3) can be reduced to cos⁡(π/3)\cos (\pi/3) which equals 1/2.
145 Medium
145 of 294
Application +4 | -1
If the solutions for θ, cos⁡pθ+cos⁡qθ=0,   p>0,   q>0\mathrm{\theta },\,\cos p\mathrm{\theta } + \cos q\mathrm{\theta } = 0,\;\,p > 0,\;\,q > 0 are in AP, then the numerically smallest common difference of AP is
View answer and explanation
Correct option: B
2πp+q\frac{{2\pi }}{{p + q}}
Option B (2πp+q\frac{{2\pi }}{{p + q}}) is correct because it represents the smallest common difference of the arithmetic progression formed by the solutions of the given trigonometric equation.
146 Medium
146 of 294
Application +4 | -1
The general value of xx for which cos⁡  2x,  12    and  sin⁡  2x\cos \,\,2x,\,\,\frac{1}{2}\;\,\,\mathrm{and}\,\,\sin \,\,2x are in AP, are given by
View answer and explanation
Correct option: B
nπ,   nπ+π4n\pi ,\;\,n\pi + \frac{\pi }{4}
Option B (-1) is correct because the sum telescopes to cos⁡π\cos \pi which equals -1.
147 Medium
147 of 294
Application +4 | -1
If 0≤x≤π20 \leq x \leq \frac{\pi}{2} and 81sin⁡2x+81cos⁡2x=3081^{\sin ^2 x}+81^{\cos ^2 x}=30, then xx is equal to
View answer and explanation
Correct option: A
π6,π3\frac{\pi}{6}, \frac{\pi}{3}
Option A (π6,π3\frac{\pi}{6}, \frac{\pi}{3}) is correct because solving the given equation by substituting y=81sin⁡2xy = 81^{\sin^2 x} leads to finding the values of sin⁡x\sin x and hence xx.
148 Medium
148 of 294
Application +4 | -1
The solution of tan⁡x+tan⁡2x+tan⁡3x=tan⁡xtan⁡2xtan⁡3x\tan x+\tan 2 x+\tan 3 x=\tan x \tan 2 x \tan 3 x is
View answer and explanation
Correct option: C
x=nπ3,n∈Ix=\frac{n \pi}{3}, n \in I
Option C (x=nπ3,n∈Ix=\frac{{n \pi}}{{3}}, n \in I) is correct because it is derived by using the tangent addition formula and simplifying the given equation.
149 Medium
149 of 294
Analyzing +4 | -1
If A+B=π3A+B=\frac{\pi}{3} and cos⁡A+cos⁡B=1\cos A+\cos B=1, then
View answer and explanation
Correct option: B
∣cos⁡A−cos⁡B∣=23|\cos A-\cos B|=\sqrt{\frac{2}{3}}
Option B (∣cos⁡A−cos⁡B∣=23|\cos A-\cos B|=\sqrt{\frac{2}{3}}) is correct because it is the result of applying trigonometric identities and the given conditions to simplify the expression.
150 Medium
150 of 294
Application +4 | -1
General solution of equation ∣sin⁡x∣=cos⁡x|\sin x|=\cos x is
View answer and explanation
Correct option: C
2nπ±π4,n∈l2 \mathrm{n} \pi \pm \frac{\pi}{4}, \mathrm{n} \in \mathrm{l}
Option C (2nπ±π4,n∈l2 \mathrm{n} \pi \pm \frac{\pi}{4}, \mathrm{n} \in \mathrm{l}) is correct because it represents the general solution considering both positive and negative cases of sin⁡x\sin x within the given condition.
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