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JEE • MATHEMATICS • TRIGONOMETRIC EQUATIONS

Trigonometric equations practice · Mixed difficulty

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Showing 301-330 of 581 questions

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301 Easy
301 of 581
Application +4 | -1
If tan⁡  (πcos⁡  θ)=cot⁡(π sin⁡θ),   0<θ<3π4\tan \,\,\left( {\pi \cos \,\,\theta } \right) = \cot \left( {\pi \,\sin \theta } \right),\;\,0 < \theta < \frac{{3\pi }}{4}, then sin⁡(θ+π4)\sin \left( {\theta + \frac{\pi }{4}} \right) equals

View answer and explanation
Correct option: C
122\frac{1}{{2\sqrt 2 }}
Option C (122\frac{1}{{2\sqrt 2 }}) is correct because solving the given equation leads to sin⁡(θ+π4)=122\sin(\theta + \frac{\pi}{4}) = \frac{1}{{2\sqrt 2 }}.
302 Easy
302 of 581
Application +4 | -1
If cot⁡xcot⁡y=k\cot x \cot y=k and x+y=π3x+y=\frac{\pi}{3}, then tan⁡x,tan⁡y\tan x, \tan y satisfy the equation
View answer and explanation
Correct option: A
kt2−3(k−1)t+1=0k t^2-\sqrt{3}(k-1) t+1=0
Option A (kt2−3(k−1)t+1=0k t^2-\sqrt{3}(k-1) t+1=0) is correct because it represents the quadratic equation whose roots are tan⁡x\tan x and tan⁡y\tan y, derived from the given conditions cot⁡xcot⁡y=k\cot x \cot y = k and x+y=π3x+y = \frac{\pi}{3}.
303 Easy
303 of 581
Application +4 | -1
Solution of the equation 3tan⁡  (θ−15)=tan⁡  (θ+15)3\tan \,\,\left( {\theta - 15} \right) = \tan \,\,\left( {\theta + 15} \right) is
View answer and explanation
Correct option: D
θ=nπ+π4\theta = n\pi + \frac{\pi }{4}
Option D (θ=nπ+π4\theta = n\pi + \frac{\pi }{4}) is correct because it represents the general solution to the given trigonometric equation, found by applying trigonometric identities and simplifying.
304 Medium
304 of 581
Analyzing +4 | -1
If αcos⁡23θ+βcos⁡4θ=16cos⁡6θ+9cos⁡2θ\alpha \cos ^2 3 \theta+\beta \cos ^4 \theta=16 \cos ^6 \theta+9 \cos ^2 \theta is an identity, then
View answer and explanation
Correct option: B
α=1,β=24\alpha=1, \beta=24
Option B (α=1,β=24\alpha=1, \beta=24) is correct because substituting these values into the given identity and simplifying leads to a valid equation.
305 Medium
305 of 581
Analyzing +4 | -1
If 0≤a≤3,0≤b≤30 \leq a \leq 3,0 \leq b \leq 3 and the equation x2+4+3cos⁡(ax+b)=2xx^2+4+3 \cos (a x+b)=2 x has atleast one solution, then the value of a+ba+b is
View answer and explanation
Correct option: C
π\pi
Option C (π\pi) is correct because the equation has at least one solution only if cos⁡(ax+b)=−1\cos(ax+b) = -1 when x=1x=1, leading to a+b=πa+b = \pi.
306 Medium
306 of 581
Application +4 | -1
If cos⁡(θ−α)=a\cos (\theta-\alpha)=a and sin⁡(θ−β)=b\sin (\theta-\beta)=b, then cos⁡2(α−β)+2absin⁡(α−β)\cos ^2(\alpha-\beta)+2 a b \sin (\alpha-\beta) is equal to
View answer and explanation
Correct option: C
a2+b2a^2+b^2
The correct answer is a2+b2a^2+b^2 because the expression simplifies to a2+b2a^2 + b^2 using trigonometric identities and the given substitutions.
307 Medium
307 of 581
Application +4 | -1
If x=cos⁡15∘x=\cos 15^{\circ}, then which of the following is correct ?
View answer and explanation
Correct option: D
16x4−16x2+1=016 x^4-16 x^2+1=0
Option D (16x4−16x2+1=016 x^4-16 x^2+1=0) is correct because it is derived from the given condition x=cos⁡15∘x = \cos 15^{\circ} by algebraic manipulation.
308 Easy
308 of 581
Analyzing +4 | -1
If cos⁡  2x=(2+1)  (cos⁡  x−12),  cos⁡  x≠12\cos \,\,2x = \left( {\sqrt 2 + 1} \right)\,\,\left( {\cos \,\,x - \frac{1}{{\sqrt 2 }}} \right),\,\,\cos \,\,x \ne \frac{1}{2}, then x∈Ix \in I
View answer and explanation
Correct option: D
{2nπ  ±  π4:n∈Z}\left\{ {2n\pi \,\, \pm \,\,\frac{\pi }{4}:n \in Z} \right\}
Option D ({2nπ±π4:n∈Z}\{ {2n\pi \pm \frac{\pi }{4}:n \in Z} \}) is correct because it represents the general solution obtained after solving the given trigonometric equation.
309 Easy
309 of 581
Application +4 | -1
The number of the solutions of the equation sin⁡5xcos⁡3x\sin 5 x \cos 3 x =sin⁡6xcos⁡2x=\sin 6 x \cos 2 x in the interval [0,π][0, \pi], is
View answer and explanation
Correct option: C
5
The correct answer is 5 because there are 5 solutions to the given equation within the specified interval [0, π\pi], found by simplifying the equation and solving for x.
310 Medium
310 of 581
Analyzing +4 | -1
In a △ABC,tan⁡A2\triangle A B C, \tan \frac{A}{2} and tan⁡B2\tan \frac{B}{2} satisfy 6x2−5x+1=06 x^2-5 x+1=0. Then
View answer and explanation
Correct option: C
a2+b2=c2a^2+b^2=c^2
Option C (a2+b2=c2a^2+b^2=c^2) is correct because the given condition implies that the triangle is right-angled at C, hence satisfying the Pythagorean theorem.
311 Medium
311 of 581
Analyzing +4 | -1
If 2  sin⁡  A2=1+sin⁡  A+1−sin⁡  A2\,\,\sin \,\,\frac{A}{2} = \sqrt {1 + \sin \,\,A} + \sqrt {1 - \sin \,\,A}, then A2\frac{A}{2} lies between
View answer and explanation
Correct option: A
2n  π+π42n\;\pi + \frac{\pi }{4} and 2n  π+3 π4,   n∈Z2n\;\pi + \frac{{3\,\pi }}{4},\;\,n \in Z
Option A (2n;π+π42n\\;\pi + \frac{\pi }{4} and 2n;π+3,π4,   n∈Z2n\\;\pi + \frac{{3\\,\pi }}{4},\;\,n \in Z) is correct because it defines the interval where A2\frac{A}{2} lies, based on the given equation and trigonometric properties.
312 Easy
312 of 581
Analyzing +4 | -1
If f(θ)=∣sin⁡θ∣+∣cos⁡θ∣,θ∈Rf(\theta)=|\sin \theta|+|\cos \theta|, \theta \in R, then
View answer and explanation
Correct option: D
f(θ)∈[1,2]f(\theta) \in[1, \sqrt{2}]
Option D (f(θ)∈[1,2]f(\theta) \in[1, \sqrt{2}]) is correct because the function's range is determined by the minimum and maximum values of ∣sin⁡θ∣+∣cos⁡θ∣|\sin \theta| + |\cos \theta|.
313 Hard
313 of 581
Analyzing +4 | -1
The solution set of the inequation log⁡1/2sin⁡x>log⁡1/2cos⁡x{\log _{1/2}}\sin x > {\log _{1/2}}\cos x in [0,   2π]\left[ {0,\;\,2\pi } \right], is

View answer and explanation
Correct option: C
(0,  π/4)\left( {0,\,\,\pi /4} \right)
Option C is correct because it correctly identifies the interval where sin x < cos x and both sin x and cos x are positive within the given range [0, 2π].
314 Easy
314 of 581
Application +4 | -1
The general value of θ\theta,obtained from the equation cos⁡2θ=sin⁡α\cos 2 \theta=\sin \alpha, is
View answer and explanation
Correct option: A
θ=nπ±(π4−α2)\theta=n \pi \pm\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)
Option A (θ=nπ±(π4−α2)\theta=n \pi \pm\left(\frac{{\pi}}{{4}}-\frac{{\alpha}}{{2}}\right)) is correct because it is derived by converting sine to cosine and applying the general solution formula.
315 Medium
315 of 581
Application +4 | -1
If D=∣1cos⁡θ1−sin⁡θ1−cos⁡θ−1sin⁡θ1∣D=\left|\begin{array}{ccc}1 & \cos \theta & 1 \\ -\sin \theta & 1 & -\cos \theta \\ -1 & \sin \theta & 1\end{array}\right|

f(x)f(x) 52θ5^2 \theta then DD lies in the interval
View answer and explanation
Correct option: C
[2−2,2+2][2-\sqrt{2}, 2+\sqrt{2}]
Option C ([2−2,2+2][2-\sqrt{2}, 2+\sqrt{2}]) is correct because by expanding the determinant and simplifying the expression, we can find the range of values for D.

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316 Easy
316 of 581
Application +4 | -1
If 1+sin⁡  θ+sin⁡2  θ+…1 + \sin \,\,\theta + {\sin ^2}\,\,\theta + \ldots to ∞=4+2 3,  0<θ<π,  θ≠π2\infty = 4 + 2\,\sqrt 3 ,\,\,0 < \theta < \pi ,\,\,\theta \ne \frac{\pi }{2}, then θ=\theta =
View answer and explanation
Correct option: D
π3    or, 2  π3\frac{\pi }{3}\;\,\,\mathrm{or},\,\frac{{2\;\pi }}{3}
Option D (π3\frac{\pi}{3} or 2π3\frac{2\pi}{3}) is correct because solving the infinite geometric series equation results in sin⁡θ=32\sin \theta = \frac{\sqrt{3}}{2}, which gives the solutions θ=π3\theta = \frac{\pi}{3} and θ=2π3\theta = \frac{2\pi}{3}.
317 Medium
317 of 581
Analyzing +4 | -1
If A,BA, B and CC are angles of anyone of the triangle such that tan⁡A+tan⁡B+tan⁡C=100\tan A+\tan B+\tan C=100. Then,
View answer and explanation
Correct option: A
There exist exactly three non-similar isosceles triangles
Option A (There exist exactly three non-similar isosceles triangles) is correct because the equation x3−50x2+50=0x^3-50 x^2+50=0 has three distinct real roots, implying three different values for angle A.
318 Easy
318 of 581
Application +4 | -1
If x=sin⁡θ∣sin⁡θ∣,y=cos⁡θ∣cos⁡θ∣x=\sin \theta|\sin \theta|, y=\cos \theta|\cos \theta|, where 99π2≤θ≤50π\frac{99 \pi}{2} \leq \theta \leq 50 \pi, then
View answer and explanation
Correct option: D
y−x=1y-x=1
Option D (y−x=1y-x=1) is correct because given the range of θ\theta, x=−sin⁡2θx = -\sin^2 \theta and y=cos⁡2θy = \cos^2 \theta, thus y−x=cos⁡2θ+sin⁡2θ=1y - x = \cos^2 \theta + \sin^2 \theta = 1.
319 Easy
319 of 581
Application +4 | -1
General solution of the equation cot⁡  θ  −  tan⁡  θ=2\cot \,\,\mathrm{\theta }\,\, - \,\,\tan \,\,\mathrm{\theta } = 2 is
View answer and explanation
Correct option: B
nπ2+π8\frac{{n\pi }}{2} + \frac{\pi }{8}
Option B (nπ2+π8\frac{{n\pi }}{2} + \frac{\pi }{8}) is correct because by simplifying the given equation using trigonometric identities, we get tan⁡2θ=1\tan 2\theta = 1, which leads to the general solution θ=nπ2+π8\theta = \frac{{n\pi }}{2} + \frac{\pi }{8}.
320 Medium
320 of 581
Analyzing +4 | -1
If x=sin⁡(α−β)sin⁡(γ−δ),  y=sin⁡(β−γ)sin⁡(α−δ)x = \sin \left( {\alpha - \beta } \right)\sin \left( {\gamma - \delta } \right),\,\,y = \sin \left( {\beta - \gamma } \right)\sin \left( {\alpha - \delta } \right) and z=sin⁡(γ−α)sin⁡(β−δ),z = \sin \left( {\gamma - \alpha } \right)\sin \left( {\beta - \delta } \right), then
View answer and explanation
Correct option: A
x+y+z=0x+y+z=0
Option A (x+y+z=0x+y+z=0) is correct because when the expressions for x, y, and z are expanded and added, the terms cancel out, resulting in zero.
321 Easy
321 of 581
Analyzing +4 | -1
The number of real solutions of sin⁡ex⋅cos⁡ex=2x−2+2−x−2\sin e^x \cdot \cos e^x=2^{x-2}+2^{-x-2} is
View answer and explanation
Correct option: A
0
Option A (0) is correct because the equation sin⁡ex⋅cos⁡ex=2x−2+2−x−2\sin e^x \cdot \cos e^x=2^{x-2}+2^{-x-2} has no real solutions since the LHS is less than or equal to 1/2 and the RHS is greater than or equal to 1/2, and equality cannot hold.
322 Easy
322 of 581
Application +4 | -1
The number of solutions of the equation 32sec⁡2x+1=10.3tan⁡2x3^{2 \sec ^2 x}+1=10.3^{\tan ^2 x} in the interval [0,2π][0,2 \pi] is
View answer and explanation
Correct option: C
4
Option C (4) is correct because solving the equation 32sec⁡2x+1=10⋅3tan⁡2x3^{2 \sec^2 x} + 1 = 10 \cdot 3^{\tan^2 x} leads to four solutions for xx in the interval [0,2π][0, 2\pi].
323 Easy
323 of 581
Application +4 | -1
The value of 2sin⁡23π4+2cos⁡2π4+2sec⁡2π32 \sin ^2 \frac{3 \pi}{4}+2 \cos ^2 \frac{\pi}{4}+2 \sec ^2 \frac{\pi}{3} is
View answer and explanation
Correct option: A
10
Option A (10) is correct because substituting the values of the trigonometric functions and simplifying the expression results in 10.
324 Easy
324 of 581
Application +4 | -1
The value of cot⁡x−tan⁡xcot⁡2x\frac{\cot x-\tan x}{\cot 2 x} is
View answer and explanation
Correct option: B
2
Option B (2) is correct because the expression simplifies to 2 using trigonometric identities for cotangent and tangent functions.
325 Easy
325 of 581
Application +4 | -1
Let S={θ∈[−2π,  2π]:2cos⁡2θ+3sin⁡θ=0}S = \left\{ {\theta \in \left[ { - 2\pi ,\,\,2\pi } \right]:2{{\cos }^2}\theta + 3\sin \theta = 0} \right\}, then the sum of the elements of SS is
View answer and explanation
Correct option: A
2π2\pi
Option A (2π2\pi) is correct because it is the sum of all solutions of the trigonometric equation within the given interval.
326 Medium
326 of 581
Analyzing +4 | -1
The equation esin⁡x−e−sin⁡x−4=0e^{\sin x}-e^{-\sin x}-4=0 has :
View answer and explanation
Correct option: B
No real roots
The equation has no real roots because the value of esin⁡xe^{\sin x} cannot be negative or less than 1, and ln⁡(2+5)>1\ln(2+\sqrt{5}) > 1, which is outside the range of sin⁡x\sin x.
327 Easy
327 of 581
Application +4 | -1
Let nn be a fixed positive integer such that sin⁡π2n+cos⁡π2n=n2\sin \frac{\pi}{2 n}+\cos \frac{\pi}{2 n}=\frac{\sqrt{n}}{2}, then
View answer and explanation
Correct option: C
n=6n=6
Option C (n=6n=6) is correct because substituting n=6n=6 into the equation sin⁡π2n+cos⁡π2n=n2\sin \frac{\pi}{2 n}+\cos \frac{\pi}{2 n}=\frac{\sqrt{n}}{2} satisfies the equation, while other options do not.
328 Medium
328 of 581
Application +4 | -1
The solution of tan⁡  2θ  tan⁡  θ=1\tan \,\,2\theta \,\,\tan \,\,\theta = 1 is 22\mathrm{22}
View answer and explanation
Correct option: B
(6n  ±  1)π6\left( {6n\,\, \pm \,\,1} \right)\frac{\pi }{6}
Option B is the correct solution because it represents the general solution to the trigonometric equation tan⁡2θtan⁡θ=1\tan 2\theta \tan \theta = 1 after simplification and solving for θ\theta.
329 Easy
329 of 581
Application +4 | -1
If sin⁡  (π4  cot⁡  θ)=cos⁡  (π4  tan⁡  θ)\sin \,\,\left( {\frac{\pi }{4}\,\,\cot \,\,\theta } \right) = \cos \,\,\left( {\frac{\pi }{4}\,\,\tan \,\,\theta } \right), then θ\theta is equal to
View answer and explanation
Correct option: D
nπ  +  π4n\pi \,\, + \,\,\frac{\pi }{4}
Option D (516\frac{5}{16}) is correct because by using trigonometric identities and known values of sine functions, the expression simplifies to 516\frac{5}{16}.
330 Easy
330 of 581
Application +4 | -1
If 1+sin⁡  x+sin⁡2  x+sin⁡3  x+…+…∞1 + \sin \,\,x + {\sin ^2}\,\,x + {\sin ^3}\,\,x + \ldots + \ldots \infty is equal to 4+23,   0<x<π4 + 2\sqrt 3 ,\,\;0 < x < \pi, then x=x =

View answer and explanation
Correct option: D
π3\frac{\pi }{3} or 2 π3\frac{{2\,\pi }}{3}
Option D (π3\frac{\pi }{3} or 2,π3\frac{{2\\,\pi }}{3}) is correct because these are the solutions to the trigonometric equation derived from the infinite geometric series.
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