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JEE • MATHEMATICS • TRIGONOMETRIC EQUATIONS

Trigonometric equations practice · Medium

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Showing 181-210 of 294 questions

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181 Medium
181 of 294
Analyzing +4 | -1
The number of distinct solution of sin⁡5θcos⁡3θ=sin⁡9θcos⁡7θ\sin 5 \theta \cos 3 \theta=\sin 9 \theta \cos 7 \theta in [0,π2]\left[0, \frac{\pi}{2}\right], is
View answer and explanation
Correct option: D
9
Option D (9) is correct because there are nine distinct solutions for θ\theta in the given interval after simplifying the trigonometric equation.
182 Medium
182 of 294
Application +4 | -1
If 5(tan⁡2x−cos⁡2x)=2cos⁡2x+95\left( {{{\tan }^2}x - {{\cos }^2}x} \right) = 2\cos 2x + 9, then the value of cos⁡ 4x\cos \,4x is
View answer and explanation
Correct option: D
−79- \frac{7}{9}
Option D (−79- \frac{7}{9}) is correct because it is the value of cos⁡4x\cos 4x calculated from the given trigonometric equation.
183 Medium
183 of 294
Analyzing +4 | -1
cos⁡2π7+cos⁡4π7+cos⁡6π7\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7}
View answer and explanation
Correct option: C
Is a negative number
Option C (Is a negative number) is correct because the sum of the given cosines is equal to -1/2, which is a negative number.
184 Medium
184 of 294
Application +4 | -1
Thesumof all solutions of the equation
cos⁡x⋅cos⁡(π3+x)⋅cos⁡(π3−x)=14,x∈[0,6π]\cos x \cdot \cos \left(\frac{\pi}{3}+x\right) \cdot \cos \left(\frac{\pi}{3}-x\right)=\frac{1}{4}, x \in[0,6 \pi] is
View answer and explanation
Correct option: B
30π30 \pi
Option B (30π30 \pi) is correct because it is the sum of all solutions of the given equation in the specified interval.
185 Medium
185 of 294
Analyzing +4 | -1
The range of 14∣sin⁡x∣−3\frac{1}{4|\sin x|-3} will be
View answer and explanation
Correct option: B
(−∞,−13]∪[1,∞)\left(-\infty,-\frac{1}{3}\right] \cup[1, \infty)
Option B ((-∞, -1/3] ∪ [1, ∞)) is correct because it represents the range of the function considering the bounds of the sine function.
186 Medium
186 of 294
Analyzing +4 | -1
The set of all xx in (−π,π)(-\pi, \pi) satisfying ∣4sin⁡x−1∣<5|4 \sin x-1| < \sqrt{5} is given by
View answer and explanation
Correct option: A
(−π10,3π10)\left(-\frac{\pi}{10}, \frac{3 \pi}{10}\right)
Option A, (−π10,3π10)\left(-\frac{{\pi}}{{10}}, \frac{{3 \pi}}{{10}}\right), is correct because it represents the interval of x values that satisfy the given inequality within the specified range.
187 Medium
187 of 294
Application +4 | -1
If 3sin⁡θ+5cos⁡θ=53 \sin \theta+5 \cos \theta=5, then the value of 5sin⁡θ−3cos⁡θ5 \sin \theta-3 \cos \theta is
View answer and explanation
Correct option: B
3
Option B (3) is correct because by using the given equation and trigonometric identities, we can find the value of the expression.
188 Medium
188 of 294
Application +4 | -1
If tan⁡2x+sec⁡x−a=0\tan ^2 x+\sec x-a=0 has atleast one solution, then the complete set of values of aa is
View answer and explanation
Correct option: D
[−1,∞)[-1, \infty)
Option D ([−1,∞)\left[-1, \infty\right)) is correct because by converting the equation to a quadratic in sec⁡x\sec x and analyzing the possible values of sec⁡x\sec x, we find that a≥−1a \geq -1 for the equation to have at least one solution.
189 Medium
189 of 294
Application +4 | -1
In a △ABC,cos⁡A+cos⁡B+cos⁡C\triangle A B C, \cos A+\cos B+\cos C is equal to
View answer and explanation
Correct option: D
1+rR1+\frac{r}{R}
Option D (1+rR1+\frac{r}{R}) is correct because by using trigonometric identities and the relationship between inradius, circumradius, and angles, the expression simplifies to 1+rR1+\frac{r}{R}.
190 Medium
190 of 294
Analyzing +4 | -1
If a,b∈[0,π]a, b \in[0, \pi] and the equation x2+4+3sin⁡(ax+b)−2x=0x^2+4+3 \sin (a x+b)-2 x=0 has atleast one solution, then the value of (a+b)(a+b) can be
View answer and explanation
Correct option: B
3π2\frac{3 \pi}{2}
Option D ((n+1)24n\frac{{(n+1)^2}}{{4 n}}) is correct because it represents the maximum value of sec⁡2(x−y)\sec ^2(x-y) derived by using the AM-GM inequality after expressing sec⁡2(x−y)\sec^2(x-y) in terms of tan⁡y\tan y.
191 Medium
191 of 294
Analyzing +4 | -1
If log⁡cos⁡xtan⁡x+log⁡sin⁡xcot⁡x=0\log _{\cos x} \tan x+\log _{\sin x} \cot x=0, then the most general solution of xx is
View answer and explanation
Correct option: B
2nπ+π4,n∈Z2 n \pi+\frac{\pi}{4}, n \in Z
Option B (2nπ+π4,n∈Z2 n \pi+\frac{\pi}{4}, n \in Z) is correct because it represents the general solution of the equation, considering the domains of the logarithmic and trigonometric functions involved.
192 Medium
192 of 294
Application +4 | -1
If 4cos⁡θ+5sin⁡θ=14 \cos \theta+5 \sin \theta=1 and α\alpha satisfy equation and α∈(−π2,π2)\alpha \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) then, value of tan⁡α\tan \alpha is
View answer and explanation
Correct option: B
10−1012\frac{\sqrt{10}-10}{12}
The correct answer is B; 10−1012\frac{{\sqrt{{10}} - 10}}{{12}}. It involves using trigonometric identities to convert the given equation into a quadratic equation in terms of tan⁡(θ/2)\tan(\theta/2), solving for tan⁡(θ/2)\tan(\theta/2), and then finding tan⁡(α)\tan(\alpha).
193 Medium
193 of 294
Analyzing +4 | -1
If 2sin⁡2((π2)cos⁡2x)=1−cos⁡(πsin⁡2x),x≠(2n+1)π2,n∈I2 \sin ^2\left(\left(\frac{\pi}{2}\right) \cos ^2 x\right)=1-\cos (\pi \sin 2 x), x \neq(2 n+1) \frac{\pi}{2}, n \in I, then cos⁡2x\cos 2 x is equal to
View answer and explanation
Correct option: B
35\frac{3}{5}
Option B (35\frac{3}{5}) is correct because it is derived by simplifying the given trigonometric equation and solving for cos⁡2x\cos 2x.
194 Medium
194 of 294
Analyzing +4 | -1
If cot⁡θ  cot⁡7θ+cot⁡θ  cot⁡4θ+cot⁡4θ  cot⁡7θ=1\cot \theta \,\,\cot 7\theta + \cot \theta \,\,\cot 4\theta + \cot 4\theta \,\,\cot 7\theta = 1 then θ=\theta =
View answer and explanation
Correct option: D
n  π12,  n∈Z\frac{{n\;\pi }}{{12}},\,\,n \in Z
Option D (n;π12,,n∈Z\frac{{n\\;\pi }}{{12}},\\,n \in Z) is correct because it represents the general solution to the given trigonometric equation.
195 Medium
195 of 294
Application +4 | -1
The most general value of θ\theta for which
sin⁡  θ−cos⁡  θ=min⁡x  ∈  R  {1,   x2−4x+6}\sin \,\,\theta - \cos \,\,\theta = \mathop {\min }\limits_{x\; \in \;R} \,\,\left\{ {1,\;\,{x^2} - 4x + 6} \right\} are given by

View answer and explanation
Correct option: B
θ=nπ  +  (−1)nπ4  +  π4,  n∈Z\theta = n\pi \,\, + \,\,{\left( { - 1} \right)^n}\frac{\pi }{4}\,\, + \,\,\frac{\pi }{4},\,\,n \in Z
Option B (θ=nπ+(−1)nπ4+π4,\n∈Z\theta = n\pi \\ + \\ {\left( { - 1} \right)^n}\frac{\pi }{4}\\ + \\ \frac{\pi }{4},\n \in Z) is correct because it is the general solution of the trigonometric equation after finding the minimum value of the quadratic expression.

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196 Medium
196 of 294
Application +4 | -1
sin⁡x⋅sin⁡(60∘−x)⋅sin⁡(60∘+x)=1/8\sin x \cdot \sin \left(60^{\circ}-x\right) \cdot \sin \left(60^{\circ}+x\right)=1 / 8 and n∈Zn \in \mathbb{Z}, then:
View answer and explanation
Correct option: B
x=nπ3+(−1)nπ18x=\frac{n \pi}{3}+(-1)^n \frac{\pi}{18}
Option B (x=nπ3+(−1)nπ18x=\frac{{n \pi}}{{3}}+(-1)^n \frac{{\pi}}{{18}}) is correct because it is derived by using the trigonometric identity and solving the resulting sine equation.
197 Medium
197 of 294
Understanding +4 | -1
The equation pcos⁡x−qsin⁡x=rp \cos x - q \sin x = r admits of a solution for xx only, if
View answer and explanation
Correct option: B
−p2+q2<r<p2+q2-\sqrt{p^2+q^2} < r < \sqrt{p^2+q^2}
The correct answer is B because the equation has a solution only if rr lies within the range of the expression, which is −p2+q2≤r≤p2+q2-\sqrt{p^2+q^2} \leq r \leq \sqrt{p^2+q^2}.
198 Medium
198 of 294
Analyzing +4 | -1
The smallest positive number pp for which the equation cos⁡(psin⁡x)=sin⁡(pcos⁡x)\cos (p \sin x)=\sin (p \cos x) has a solution x∈[0,2π]x \in[0,2 \pi], is equal to
View answer and explanation
Correct option: D
π22\frac{\pi}{2 \sqrt{2}}
Option D (π22\frac{\pi}{2 \sqrt{2}}) is correct because it is the smallest positive value of pp that satisfies the given equation.
199 Medium
199 of 294
Application +4 | -1
If α+β=π2\alpha+\beta=\frac{\pi}{2} and β+γ=α\beta+\gamma=\alpha, then tan⁡α\tan \alpha is equal to
View answer and explanation
Correct option: C
tan⁡β+2tan⁡γ\tan \beta+2 \tan \gamma
Option C (tan⁡β+2tan⁡γ\tan \beta+2 \tan \gamma) is correct because it expresses tan⁡α\tan \alpha in terms of tan⁡β\tan \beta and tan⁡γ\tan \gamma based on the given relationships between the angles.
200 Medium
200 of 294
Analyzing +4 | -1
The set of values of xx for which tan⁡  3  x−tan⁡  2  x1+tan⁡  3  x  tan⁡  2  x=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\mathrm{\Phi }
Option A (Φ\Phi) is correct because the solution obtained from simplifying the equation makes the original equation undefined, thus there is no solution.
201 Medium
201 of 294
Application +4 | -1
If 2cos⁡x<32 \cos x < \sqrt{3} and x∈[−π,π]x \in[-\pi, \pi], then the solution set for xx is
View answer and explanation
Correct option: A
[−π,−π6)∪(π6,π]\left[-\pi,-\frac{\pi}{6}\right) \cup\left(\frac{\pi}{6}, \pi\right]
Option A ([-π,-π6\frac{π}{6}) ∪ (π6\frac{π}{6}, π]) is correct because it represents the interval where cos⁡x<32\cos x < \frac{{\sqrt{{3}}}}{{2}} within the given range of x∈[−π,π]x \in [-\pi, \pi].
202 Medium
202 of 294
Analyzing +4 | -1
Solution of the equation cos⁡2(12px)+cos⁡2(12qx)=1{\cos ^2}\left( {\frac{1}{2}px} \right) + {\cos ^2}\left( {\frac{1}{2}qx} \right) = 1 form an arithmetic progression with common difference
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the solution forms two arithmetic progressions with different common differences, neither of which matches the provided options.
203 Medium
203 of 294
Analyzing +4 | -1
Statement - 1: tan⁡3x−tan⁡2x1+tan⁡3xtan⁡2x=1,x=nπ+π4,n∈I\frac{\tan 3 x-\tan 2 x}{1+\tan 3 x \tan 2 x}=1, x=n \pi+\frac{\pi}{4}, n \in I
Statement - 2: tan⁡x\tan {x} is not defined at x=nπ+π2,n∈Ix=n \pi+\frac{\pi}{2}, n \in I
View answer and explanation
Correct option: D
Statement-1 is False, Statement-2 is True
Statement-1 is False, Statement-2 is True because the given equation simplifies to tan(x) = 1, but tan(2x) is undefined at x = nπ + π/4.
204 Medium
204 of 294
Application +4 | -1
The general solution for α\alpha is, if 1/6sin⁡α,cos⁡α,tan⁡α1 / 6 \sin \alpha, \cos \alpha, \tan \alpha in G.P.
View answer and explanation
Correct option: C
2nπ±π32 n \pi \pm \frac{\pi}{3}
Option C (2nπ±π32 n \pi \pm \frac{{\pi}}{{3}}) is correct because it represents the general solution derived from the given geometric progression condition and solving for α\alpha.
205 Medium
205 of 294
Analyzing +4 | -1
The number of elements in the set S={θ∈[0,2π]:3cos⁡4θ−5cos⁡2θ−2sin⁡6θ+2=0}S=\left\{\theta \in[0,2 \pi]: 3 \cos ^4 \theta-5 \cos ^2 \theta-2 \sin ^6\theta+2=0\right\} is
View answer and explanation
Correct option: D
9
Option C (Both the statements (A) and (B) are true) is correct because the given inequality implies that f(x) is bounded between 1/8 and 1 for all x in the interval [2,4].
206 Medium
206 of 294
Application +4 | -1
If a=cos⁡ϕcos⁡ψ+sin⁡ϕsin⁡ψcos⁡δa=\cos \phi \cos \psi+\sin \phi \sin \psi \cos \delta

b=cos⁡ϕsin⁡ψ−sin⁡ϕcos⁡ψcos⁡δb=\cos \phi \sin \psi-\sin \phi \cos \psi \cos \delta

and c=sin⁡ϕsin⁡δc=\sin \phi \sin \delta. Then a2+b2+c2=a^2+b^2+c^2=
View answer and explanation
Correct option: C
1
Option C (1) is correct because by expanding the expression a2 + b2 + c2 and using trigonometric identities, it simplifies to 1.
207 Medium
207 of 294
Analyzing +4 | -1
The number of solutions of the equation 16(sin⁡5x+cos⁡5x)=11(sin⁡x+cos⁡x)16\left(\sin ^5 x+\cos ^5 x\right)=11(\sin x+\cos x) in the interval [0,2π][0,2 \pi] is
View answer and explanation
Correct option: A
6
Option A (6) is correct because the equation simplifies to a trigonometric equation with 6 solutions in the given interval.
208 Medium
208 of 294
Application +4 | -1
If π2<α<π,π<β<3π2;sin⁡α=1517\frac{\pi}{2} < \alpha < \pi, \pi < \beta < \frac{3 \pi}{2} ; \sin \alpha=\frac{15}{17} and tan⁡β=125\tan \beta=\frac{12}{5}, then the value of sin⁡(β−α)\sin (\beta-\alpha) is
View answer and explanation
Correct option: D
171221\frac{171}{221}
The correct answer is 171221\frac{171}{221} because it is the result of applying the sine subtraction formula and using the given information to find the values of sine and cosine.
209 Medium
209 of 294
Analyzing +4 | -1
If sin⁡2x+cos⁡2y=2sec⁡2z\sin ^2 x+\cos ^2 y=2 \sec ^2 z, then
View answer and explanation
Correct option: A
x=(2m+1)π2,y=nπx=(2 m+1) \frac{\pi}{2}, y=n \pi and z=tπz=t \pi
Option A (x=(2m+1)π2,y=nπx=(2 m+1) \frac{\pi}{2}, y=n \pi and z=tπz=t \pi) is correct because it satisfies the given equation sin⁡2x+cos⁡2y=2sec⁡2z\sin ^2 x+\cos ^2 y=2 \sec ^2 z when each term reaches its extreme value.
210 Medium
210 of 294
Application +4 | -1
Suppose θ∈[0,π4]\theta \in\left[0, \frac{\pi}{4}\right] is a solution of 4cos⁡θ−3sin⁡θ=1.4 \cos \theta-3 \sin \theta=1. Then cos⁡θ\cos \theta is equal to:
View answer and explanation
Correct option: A
4(36−2)\frac{4}{(3 \sqrt{6}-2)}
The correct answer is D (4(36−2)\frac{{4}}{{(3 \sqrt{{6}}-2)}}) because it is the value of cos(theta) obtained by solving the trigonometric equation and using the half-angle substitution.
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