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JEE • MATHEMATICS • TRIGONOMETRIC EQUATIONS

Trigonometric equations practice · Mixed difficulty

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Showing 181-210 of 581 questions

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181 Medium
181 of 581
Analyzing +4 | -1
The different between the greatest and least values of the function f(x)=cos⁡x12cos⁡2x−13cos⁡3x  f\left( x \right) = \cos x\frac{1}{2}\cos 2x - \frac{1}{3}\cos 3x\; is
View answer and explanation
Correct option: D
94\frac{9}{4}
Option D (94\frac{9}{4}) is correct because it represents the difference between the greatest and least values of the function, calculated by finding critical points and evaluating the function at those points.
182 Medium
182 of 581
Understanding +4 | -1
The equation sin⁡x+sin⁡y+sin⁡z=−3\sin x+\sin y+\sin z=-3 for 0≤x≤2π,0≤y≤2π,0≤z≤2π0 \leq x \leq 2 \pi, 0 \leq y \leq 2 \pi, 0 \leq z \leq 2 \pi, has
View answer and explanation
Correct option: A
One solution
Option A (One solution) is correct because the equation is only satisfied when each sine term is -1, leading to a unique solution.
183 Medium
183 of 581
Analyzing +4 | -1
Let S={0∈(0,π2):∑m=19sec⁡(θ+(m−1)π6)sec⁡(θ+mπ6)=−83}.S = \left\{ {0 \in \left( {0,\frac{\pi }{2}} \right):\sum\limits_{m = 1}^9 {\sec \left( {\theta + \left( {m - 1} \right)\frac{\pi }{6}} \right)\sec \left( {\theta + \frac{{m\pi}}{6}} \right) = - \frac{8}{{\sqrt 3 }}} } \right\}. Then
View answer and explanation
Correct option: C
∑θ∈Sθ=π2\sum\limits_{\theta \in S} {\theta = \frac{\pi }{2}}
Option C (∑θ∈Sθ=π2\sum\limits_{\theta \in S} {\theta = \frac{\pi }{2}}) is correct because solving the trigonometric equation leads to two solutions for θ\theta, and their sum is π2\frac{\pi}{2}.
184 Medium
184 of 581
Analyzing +4 | -1
The number of solutions of sin⁡7x+cos⁡7x=1,x∈[0,4π]\sin ^7 x+\cos ^7 x=1, x \in[0,4 \pi] is
View answer and explanation
Correct option: C
5
Option C (5) is correct because we need to find the values of x in the interval [0,4π][0, 4\pi] for which sin⁡7x+cos⁡7x=1\sin^7 x + \cos^7 x = 1, which occurs when either sin⁡x=0\sin x = 0 and cos⁡x=1\cos x = 1 or sin⁡x=1\sin x = 1 and cos⁡x=0\cos x = 0.
185 Medium
185 of 581
Analyzing +4 | -1
If sin⁡  θ−cos⁡  θ<0\sin \,\,\theta - \cos \,\,\theta < 0, then θ\theta lies between
View answer and explanation
Correct option: D
2n  π−3π42n\;\pi - \frac{{3\pi }}{4} and 2n  π+π4,   n∈Z2n\;\pi + \frac{\pi }{4},\;\,n \in Z
Option A is correct because it represents the range of θ where sin(θ) - cos(θ) < 0, derived by transforming the inequality and solving for θ.
186 Easy
186 of 581
Application +4 | -1
If log⁡3(sin⁡x+22cos⁡x)≥2,−2π≤x≤2π{\log _{\sqrt 3 }}(\sin x + 2\sqrt 2 \cos x) \ge 2, - 2\pi \le x \le 2\pi, then the number of solutions of xx is
View answer and explanation
Correct option: D
4
Option D (4) is correct because there are 4 solutions for x in the given interval that satisfy the logarithmic inequality.
187 Medium
187 of 581
Application +4 | -1
If 0≤θ≤π20 \leq \theta \leq \frac{\pi}{2} and x=Xcos⁡θ+Ysin⁡θx=X \cos \theta+Y \sin \theta, y=Xsin⁡θ−Ycos⁡θy=X \sin \theta-Y \cos \theta such that x2+2xy+y2=aX2+bY2x^2+2 x y+y^2=a X^2+b Y^2, where aa and bb are constants, then
View answer and explanation
Correct option: C
a=3,b=−1a=3, b=-1
Option C (a=3,b=−1a=3, b=-1) is correct because by substituting the given expressions for x and y into the equation and comparing coefficients, we find that a=3a=3 and b=−1b=-1 when θ=π4\theta = \frac{{\pi}}{{4}}.
188 Medium
188 of 581
Analyzing +4 | -1
The value of 2cos⁡π13cos⁡9π13+cos⁡3π13+cos⁡5π132 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13} is
View answer and explanation
Correct option: B
0
Option B (0) is correct because the expression simplifies to zero using trigonometric identities.
189 Easy
189 of 581
Application +4 | -1
The equation 3cos⁡x+4sin⁡x=63 \cos x+4 \sin x=6 has
View answer and explanation
Correct option: D
No solution
Option D (No solution) is correct because the maximum value of 3cos⁡x+4sin⁡x3 \cos x + 4 \sin x is 5, which is less than 6.
190 Hard
190 of 581
Application +4 | -1
The ordered pair (x,y)(x, y) satisfying 1−2x−x2=tan⁡2(x+y)+cot⁡2(x+y)1-2 x-x^2=\tan ^2(x+y)+\cot ^2(x+y) is
View answer and explanation
Correct option: C
(−1,(2n+1)π4+1)\left(-1,(2 n+1) \frac{\pi}{4}+1\right)
Option C ((-1,(2n+1)π4\frac{π}{4}+1)) is correct because the equation holds true only when both sides are equal to 2, which occurs when x = -1 and y = (2n+1)π4\frac{π}{4}+1.
191 Easy
191 of 581
Application +4 | -1
If sin⁡  6  θ+sin⁡  4  θ+sin⁡  2θ=0\sin \,\,6\;\theta + \sin \,\,4\;\theta + \sin \,\,2\theta = 0 then the general value of θ   is\theta \;\,is

View answer and explanation
Correct option: A
nπ4,   nπ  ±  π3\frac{{n\pi }}{4},\,\;n\pi \,\, \pm \,\,\frac{\pi }{3}
Option A is correct because by using the sum-to-product trigonometric identities and solving the resulting equations, we find the general value of θ\theta.
192 Easy
192 of 581
Analyzing +4 | -1
If in a triangle ABC,  sin⁡A=sin⁡2BABC,\,\,\sin A = {\sin ^2}B and 2cos⁡2A=cos⁡2B2{\cos ^2}A = {\cos ^2}B then the Δ ABC\Delta \,ABC is
View answer and explanation
Correct option: B
Obtuse angled
The correct answer is Obtuse angled because the given conditions lead to angles A and B such that one of the angles in the triangle is greater than 90 degrees.
193 Hard
193 of 581
Analyzing +4 | -1
cos⁡  2x+k sin    x=2k−7\cos \,\,2x + k\,sin\,\,\;x = 2k - 7 has a solution for
View answer and explanation
Correct option: A
2≤k≤62 \le k \le 6
Option A (2≤k≤62 \le k \le 6) is correct because this range of kk ensures that the value of sin⁡x\sin x obtained from the equation lies between -1 and 1, which is a necessary condition for a solution to exist.
194 Medium
194 of 581
Analyzing +4 | -1
If α\alpha and β\beta satisfy the equation 12sin⁡α+5cos⁡α12 \sin \alpha+5 \cos \alpha =2β2−8β+21=2 \beta^2-8 \beta+21, then the value of β2−2β\beta^2-2 \beta is _____.
View answer and explanation
Correct option: A
0
Option A (0) is correct because the maximum value of 12sin⁡α+5cos⁡α12 \sin \alpha + 5 \cos \alpha is 13, and the minimum value of 2β2−8β+212 \beta^2 - 8 \beta + 21 is also 13, which occurs when β=2\beta = 2. Therefore, β2−2β=0\beta^2 - 2\beta = 0.
195 Medium
195 of 581
Application +4 | -1
If 2sec⁡θ+tan⁡θ=1\sqrt{2} \sec \theta+\tan \theta=1, then the general value θ\theta is
View answer and explanation
Correct option: C
2nπ−π42 n \pi-\frac{\pi}{4}
Option C (2nπ−π42 n \pi-\frac{{\pi}}{{4}}) is correct because it represents the general solution derived from solving the given trigonometric equation.

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196 Medium
196 of 581
Analyzing +4 | -1
The value of the expression sin⁡3x1+cos⁡x+cos⁡3x1−sin⁡x\frac{\sin ^3 x}{1+\cos x}+\frac{\cos ^3 x}{1-\sin x} is
View answer and explanation
Correct option: A
2cos⁡[π4−x]\sqrt{2} \cos \left[\frac{\pi}{4}-x\right]
Option A (2cos⁡[π4−x]\sqrt{2} \cos \left[\frac{\pi}{4}-x\right]) is correct because the expression simplifies to sin⁡x+cos⁡x\sin x + \cos x, which is equivalent to 2cos⁡[π4−x]\sqrt{2} \cos \left[\frac{\pi}{4}-x\right].
197 Easy
197 of 581
Application +4 | -1
The solution set of (2cos⁡x−1)(3+2cos⁡x)=0(2 \cos x-1)(3+2 \cos x)=0 in the interval 0≤x≤2π0 \leq x \leq 2 \pi is
View answer and explanation
Correct option: B
{π3,5π3}\left\{\frac{\pi}{3}, \frac{5 \pi}{3}\right\}
Option B ({π3,5π3}\{\frac{\pi}{3}, \frac{5 \pi}{3}\}) is correct because these are the only solutions to the equation (2cos⁡x−1)(3+2cos⁡x)=0(2 \cos x-1)(3+2 \cos x)=0 within the given interval 0≤x≤2π0 \leq x \leq 2 \pi.
198 Medium
198 of 581
Application +4 | -1
If 2  sin⁡2θ=3  cos⁡  θ,  0≤θ≤2π2\,\,{\sin ^2}\mathrm{\theta } = 3\,\,\cos \,\,\mathrm{\theta },\mathrm{\;}0 \le \mathrm{\theta } \le 2\mathrm{\pi } then θ\mathrm{\theta } is equal to

View answer and explanation
Correct option: C
π3,  5 π3\frac{\mathrm{\pi }}{3},\,\,\frac{{5\,\mathrm{\pi }}}{3}
Option C (π3,  5 π3\frac{\mathrm{\pi }}{3},\,\,\frac{{5\,\mathrm{\pi }}}{3}) is correct because these values of θ\theta satisfy the given trigonometric equation within the specified range of 0≤θ≤2π0 \le \mathrm{\theta } \le 2\mathrm{\pi }.
199 Medium
199 of 581
Analyzing +4 | -1
If cos⁡α+2cos⁡β+3cos⁡γ=0,sin⁡α+2sin⁡β\cos \alpha+2 \cos \beta+3 \cos \gamma=0, \sin \alpha+2 \sin \beta +3sin⁡γ=0+3 \sin \gamma=0 and α+β+γ=π\alpha+\beta+\gamma=\pi, then sin⁡3α+8sin⁡3β+27sin⁡3γ\sin 3 \alpha+8 \sin 3 \beta+27 \sin 3 \gamma is equal to
View answer and explanation
Correct option: B
0
Option B is correct because using complex numbers and the given conditions, it can be shown that sin⁡3α+8sin⁡3β+27sin⁡3γ=0\sin 3 \alpha+8 \sin 3 \beta+27 \sin 3 \gamma = 0.
200 Easy
200 of 581
Application +4 | -1
If tan⁡θ+sec⁡θ=3,0<θ<π\tan \theta+\sec \theta=\sqrt{3}, 0 < \theta < \pi, then θ\theta is equal to
View answer and explanation
Correct option: C
π/6\pi / 6
Option C (π/6\pi / 6) is correct because by rewriting the equation using trigonometric identities and solving for theta, we find that theta = pi/6.
201 Medium
201 of 581
Analyzing +4 | -1
The number of solutions of the equation (1−cos⁡2x)(1-\cos 2 x) (cos⁡2x+cos⁡2x)=0,0≤x≤2π\left(\cos 2 x+\cos ^2 x\right)=0,0 \leq x \leq 2 \pi is
View answer and explanation
Correct option: D
0
Option D (0) is correct because the given equation has no solutions in the specified interval due to the cotangent function being undefined.
202 Easy
202 of 581
Application +4 | -1
The equation cos⁡2x+asin⁡x=2a−7\cos 2 x+a \sin x=2 a-7 possesses a solution if
View answer and explanation
Correct option: B
2≤a≤62 \leq a \leq 6
Option B (2≤a≤62 \leq a \leq 6) is correct because the equation has a solution only when the value of sin⁡x\sin x derived from the quadratic equation lies between -1 and 1.
203 Medium
203 of 581
Application +4 | -1
If tan⁡α=ba,a>b>0\tan \alpha=\frac{b}{a}, a > b > 0 and 0<α<π40 < \alpha < \frac{\pi}{4}, then a+ba−b−a−ba+b\sqrt{\frac{a+b}{a-b}}-\sqrt{\frac{a-b}{a+b}} is equal to
View answer and explanation
Correct option: A
2sin⁡αcos⁡2α\frac{2 \sin \alpha}{\sqrt{\cos 2 \alpha}}
Option A (2sin⁡αcos⁡2α\frac{{2 \sin \alpha}}{{\sqrt{{\cos 2 \alpha}}}}) is correct because it is derived by simplifying the given expression using trigonometric identities and the given condition tan⁡α=ba\tan \alpha = \frac{{b}}{{a}}.
204 Easy
204 of 581
Application +4 | -1
If cos⁡θ+cos⁡7θ+cos⁡3θ+cos⁡5θ=0\cos \theta+\cos 7 \theta+\cos 3 \theta+\cos 5 \theta=0, then θ=\theta=
View answer and explanation
Correct option: C
nπ8\frac{n \pi}{8}
Option C (nπ8\frac{{n \pi}}{{8}}) is correct because solving the trigonometric equation leads to sin⁡8θ=0\sin 8 \theta = 0, thus θ=nπ8\theta = \frac{{n \pi}}{{8}}.
205 Medium
205 of 581
Application +4 | -1
The most general solutions of the equation sec⁡2x=2(1−tan⁡2x)\sec ^2 x=\sqrt{2}\left(1-\tan ^2 x\right) are given by
View answer and explanation
Correct option: C
nπ±π8n \pi \pm \frac{\pi}{8}
Option C (nπ±π8n \pi \pm \frac{\pi}{8}) is correct because by simplifying the given equation, we get tan⁡2x=tan⁡2π8\tan^2 x = \tan^2 \frac{\pi}{8}, which leads to the general solution x=nπ±π8x = n \pi \pm \frac{\pi}{8}.
206 Hard
206 of 581
Analyzing +4 | -1
If the equation cos⁡  (λ  sin⁡  θ)=sin⁡(λ  cos⁡  θ)\cos \,\,\left( {\lambda \,\,\sin \,\,\theta } \right) = \sin \left( {\lambda \,\,\cos \,\,\theta } \right) has a solution in [0,   2  π]\left[ {0,\;\,2\;\pi } \right], then the smallest positive value of λ\lambda is
View answer and explanation
Correct option: D
π2 2\frac{\pi }{{2\,\sqrt 2 }}
Option D (π2,2\frac{\pi }{{2\\,\sqrt 2 }}) is correct because it represents the smallest positive value of λ\lambda that allows the given trigonometric equation to have a solution within the specified interval.
207 Medium
207 of 581
Application +4 | -1
If sin⁡α,1\sin \alpha, 1 and cos⁡2α\cos 2 \alpha are in GP, then α\alpha is equal to
View answer and explanation
Correct option: B
nπ+(−1)n−1π2,n∈In \pi+(-1)^{n-1} \frac{\pi}{2}, n \in I
Option B (nπ+(−1)n−1π2,n∈In \pi+(-1)^{n-1} \frac{\pi}{2}, n \in I) is correct because when sin⁡α,1\sin \alpha, 1 and cos⁡2α\cos 2 \alpha are in GP, the only possible values are when both sin⁡α\sin \alpha and cos⁡2α\cos 2 \alpha are 1 or -1, leading to α=nπ+(−1)n−1π2\alpha = n \pi+(-1)^{n-1} \frac{\pi}{2}.
208 Medium
208 of 581
Application +4 | -1
If 2sin⁡2(x−π3)−5sin⁡(x−π3)+2<02 \sin ^2\left(x-\frac{\pi}{3}\right)-5 \sin \left(x-\frac{\pi}{3}\right)+2 < 0, then xx belongs to-
View answer and explanation
Correct option: D
((4n+1)π2,(12n+7)π6),n∈I\left((4 n+1) \frac{\pi}{2},(12 n+7) \frac{\pi}{6}\right), n \in I
Option D (((4n+1)π2,(12n+7)π6),n∈I((4 n+1) \frac{\pi}{2},(12 n+7) \frac{\pi}{6}), n \in I) is correct because it represents the solution set for x after solving the given trigonometric inequality.
209 Medium
209 of 581
Analyzing +4 | -1
Let A={θ:sin⁡(θ)=tan⁡(θ)}{A}=\{\theta: \sin (\theta)=\tan (\theta)\} and B={θ:cos⁡(θ)=1}{B}=\{\theta: \cos (\theta)=1\} be two sets. Then :
View answer and explanation
Correct option: B
A⊄BA \not \subset B
The correct answer is A ⊄B\not\subset B because set A contains elements that are not in set B, specifically odd multiples of \π\pi.
210 Medium
210 of 581
Analyzing +4 | -1
If sin⁡θ=3sin⁡(θ+2α)\sin \theta=3 \sin (\theta+2 \alpha), then the value of tan⁡(θ+α)+2tan⁡α\tan (\theta+\alpha)+2 \tan \alpha is
View answer and explanation
Correct option: D
0
Option D (0) is correct because applying componendo and dividendo and simplifying leads to the given expression equaling zero.
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