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JEE • MATHEMATICS • TRIGONOMETRIC EQUATIONS

Trigonometric equations practice · Medium

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211 Medium
211 of 294
Application +4 | -1
The value of ∑r=010cos⁡3πr3\sum\limits_{{r=0}}^{10} \cos ^3 \frac{\pi r}{3} is
View answer and explanation
Correct option: D
−18-\frac{1}{8}
Option D (-18\frac{1}{8}) is correct because by using the trigonometric identity cos⁡3x=14(cos⁡3x+3cos⁡x)\cos^3 x = \frac{1}{4}(\cos 3x + 3 \cos x) and simplifying the summation, the value is found to be -18\frac{1}{8}.
212 Medium
212 of 294
Analyzing +4 | -1
If 3sin⁡2θ+2sin⁡2ϕ=13 \sin ^2 \theta+2 \sin ^2 \phi=1 and 3sin⁡2θ=2sin⁡2ϕ,0<θ<π23 \sin 2 \theta=2 \sin 2 \phi, 0 < \theta < \frac{\pi}{2} and 0<ϕ<π20 < \phi < \frac{\pi}{2}, then the value of θ+2ϕ\theta+2 \phi is
View answer and explanation
Correct option: A
π2\frac{\pi}{2}
Option A (π2\frac{\pi}{2}) is correct because after simplification and using trigonometric identities, θ+2ϕ\theta + 2\phi equals π2\frac{\pi}{2}.
213 Medium
213 of 294
Analyzing +4 | -1
The number of solution of 2sin⁡x=3x+1+31−x2 \sin x=3^{x+1}+3^{1-x} is
View answer and explanation
Correct option: C
Zero solution
Option C (Zero solution) is correct because the left-hand side is bounded by 2, while the right-hand side is always greater than or equal to 6.
214 Medium
214 of 294
Analyzing +4 | -1
If f(θ)=5cos⁡θ+3cos⁡(θ+π3)+3f(\theta)=5 \cos \theta+3 \cos \left(\theta+\frac{\pi}{3}\right)+3, then range of f(θ)f(\theta) is
View answer and explanation
Correct option: D
[−4,10][-4,10]
Option D ([-4,10]) is correct because it represents the range of the given trigonometric function after simplification and applying the range formula.
215 Medium
215 of 294
Application +4 | -1
The equation (cos⁡ p−1)x2+(cos⁡ p)x+sin⁡p=0\left( {\cos \,p - 1} \right){x^2} + \left( {\cos \,p} \right)x + \sin p = 0, where xx is a variable, has real roots. Then, the interval of pp may be any one of the following:
View answer and explanation
Correct option: D
(0,   π)\left( {0,\;\,\pi } \right)
Option D ((0,  π)\left( {0,\;\pi } \right)) is correct because the condition for real roots of the quadratic equation is satisfied within this interval.
216 Medium
216 of 294
Analyzing +4 | -1
If exp⁡{(sin⁡2x+sin⁡4x+sin⁡6x+…\exp \left\{\left(\sin ^2 x+\sin ^4 x+\sin ^6 x+\ldots\right.\right. upto ∞)ln⁡2}\left.\left.\infty\right) \ln 2\right\} satisfies the equation x2−17x+16=0x^2-17 x+16=0 then value of 2cos⁡xsin⁡x+2cos⁡x(0<x<π/2)\frac{2 \cos x}{\sin x+2 \cos x}(0 < x < \pi / 2) is
View answer and explanation
Correct option: A
12\frac{1}{2}
The correct answer is 12\frac{1}{2} because solving the quadratic equation gives tan⁡2x=4\tan^2 x = 4, leading to tan⁡x=2\tan x = 2, and substituting this into the expression yields 12\frac{1}{2}.
217 Medium
217 of 294
Application +4 | -1
If sin⁡  5  x+sin⁡  3  x+sin⁡  x=0\sin \,\,5\;x + \sin \,\,3\;x + \sin \,\,x = 0 then the value of xx other than zero, lying between 0≤x≤π20 \le x \le \frac{\pi }{2} is
View answer and explanation
Correct option: C
π3\frac{\pi }{3}
Option C is correct because the value of x is found by solving the trigonometric equation using sum-to-product formulas.
218 Medium
218 of 294
Application +4 | -1
The general solution for α\alpha is, if sin⁡α,1,cos⁡2α\sin \alpha, 1, \cos 2 \alpha in G.P.
View answer and explanation
Correct option: A
nπ+(−1)n+1π2n \pi+(-1)^{n+1} \frac{\pi}{2}
Option A (nπ+(−1)n+1π2n \pi+(-1)^{{n+1}} \frac{{\pi}}{{2}}) is correct because it represents the general solution derived from the given geometric progression condition and solving for α\alpha.
219 Medium
219 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\frac{\pi}{6}
Option A (π6\frac{{\pi}}{{6}}) is correct because solving the equation 81sin⁡2x+81cos⁡2x=3081^{\sin ^2 x}+81^{\cos ^2 x}=30 leads to sin⁡x=12\sin x = \frac{{1}}{{2}} or sin⁡x=32\sin x = \frac{{\sqrt{{3}}}}{{2}}, and considering the given range, x=π6x = \frac{{\pi}}{{6}} is a valid solution.
220 Medium
220 of 294
Application +4 | -1
If the complex numbers sin⁡  x+  i  cos⁡  2  x\sin \,\,x + \,\,i\,\,\cos \,\,2\;x and cos⁡  x−i  sin⁡  2  x\cos \,\,x - i\,\,\sin \,\,2\;x are conjugate to each other, then xx is equal to
View answer and explanation
Correct option: D
None of these
Option D is correct because no value of x satisfies the condition that the given complex numbers are conjugates of each other.
221 Medium
221 of 294
Application +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: B
x∈(−π10,3π10)x \in\left(-\frac{\pi}{10}, \frac{3 \pi}{10}\right)
Option B (x∈(−π10,3π10)x \in(-\frac{\pi}{10}, \frac{3 \pi}{10})) is correct because solving the inequality ∣4sin⁡x−1∣<5|4 \sin x-1| < \sqrt{5} within the interval (−π,π)(-\pi, \pi) leads to this solution set.
222 Medium
222 of 294
Application +4 | -1
The expression
(cos⁡A+cos⁡Bsin⁡A−sin⁡B)n+(sin⁡A+sin⁡Bcos⁡A−cos⁡B)n=\left(\frac{\cos \mathrm{A}+\cos \mathrm{B}}{\sin \mathrm{A}-\sin \mathrm{B}}\right)^{\mathrm{n}}+\left(\frac{\sin \mathrm{A}+\sin \mathrm{B}}{\cos \mathrm{A}-\cos \mathrm{B}}\right)^{\mathrm{n}}=
View answer and explanation
Correct option: A
2cot⁡n(A−B2)2 \cot ^{\mathrm{n}}\left(\frac{\mathrm{A}-\mathrm{B}}{2}\right) if n\mathrm{n} is even
Option A is correct because the expression simplifies to a trigonometric form, and the result depends on whether n is even or odd.
223 Medium
223 of 294
Understanding +4 | -1
sin⁡πn+sin⁡3πn+sin⁡5πn+…\sin \frac{\pi}{n}+\sin \frac{3 \pi}{n}+\sin \frac{5 \pi}{n}+\ldots to nn terms is equal to
View answer and explanation
Correct option: D
0
Option D is correct because the sum of the sine series is 0.
224 Medium
224 of 294
Application +4 | -1
Solve the equations for xx and y:y:

xcos⁡3y+3xcos⁡ysin⁡2y=14xsin⁡3y+3xcos⁡2ysin⁡y=13\begin{aligned} & x \cos ^3 y+3 x \cos y \sin ^2 y=14 \\ & x \sin ^3 y+3 x \cos ^2 y \sin y=13 \end{aligned}
View answer and explanation
Correct option: C
Both
Option C (Both) is correct because by dividing and using componendo and dividendo, we find two possible solutions for x and y that satisfy the given conditions.
225 Medium
225 of 294
Analyzing +4 | -1
If f(x)=∣sin⁡2θcos⁡2θxcos⁡2θxsin⁡2θxsin⁡2θcos⁡2θ∣,θ∈(0,π/2)f(x)=\left|\begin{array}{ccc}\sin ^2 \theta & \cos ^2 \theta & x \\ \cos ^2 \theta & x & \sin ^2 \theta \\ x & \sin ^2 \theta & \cos ^2 \theta\end{array}\right|, \theta \in(0, \pi / 2), then roots of f(x)=0f(x)=0 are
View answer and explanation
Correct option: B
1/2,−11 / 2,-1
Option B (1/2,−11/2, -1) is the correct answer because these are the values of x that make the determinant of the given matrix equal to zero.

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226 Medium
226 of 294
Application +4 | -1
The value of θ\theta satisfying cos⁡θ+3sin⁡θ=2\cos \theta+\sqrt{3} \sin \theta=2 is
View answer and explanation
Correct option: B
60∘60^{\circ}
Option B (60∘60^{\circ}) is correct because by rewriting the equation and using the cosine subtraction formula, we find that θ=60∘\theta = 60^{\circ}.
227 Medium
227 of 294
Analyzing +4 | -1
Let f:R→Rf: R \rightarrow R be a function defined by f(x)=log⁡m{2(sin⁡x−cos⁡x)+m−2}f(x)=\log _{\sqrt{m}}\{\sqrt{2}(\sin x-\cos x)+m-2\}, for some mm, such that the range of ff is [0,2][0,2]. Then the value of mm is
View answer and explanation
Correct option: D
5
Option D (5) is correct because the range of the function is [0,2], which leads to the equation log⁡m(m−4)=0\log_{\sqrt{m}}(m-4) = 0, and solving for m gives m = 5.
228 Medium
228 of 294
Analyzing +4 | -1
tan⁡81∘−tan⁡63∘−tan⁡27∘+tan⁡9∘\tan 81^{\circ}-\tan 63^{\circ}-\tan 27^{\circ}+\tan 9^{\circ} equals
View answer and explanation
Correct option: D
4
Option D (4) is correct because simplifying the expression using trigonometric identities leads to the value 4.
229 Medium
229 of 294
Analyzing +4 | -1
If cos⁡x=tan⁡y,cot⁡y=tan⁡z\cos x=\tan y, \cot y=\tan z and cot⁡z=tan⁡x\cot z=\tan x, then sin⁡x\sin x is equal to
View answer and explanation
Correct option: D
5−12\frac{\sqrt{5}-1}{2}
Option D (5−12\frac{\sqrt{5}-1}{2}) is correct because substituting the given relations leads to a quadratic equation in sin⁡x\sin x, and solving it yields this value as the valid solution.
230 Medium
230 of 294
Application +4 | -1
The number of solutions of 16sin⁡2x+16cos⁡2x=10,0≤x≤2π16^{\sin ^2 x}+16^{\cos ^2 x}=10,0 \leq x \leq 2 \pi is
View answer and explanation
Correct option: A
8
Option A (8) is correct because solving the equation 16sin⁡2x+16cos⁡2x=1016^{\sin^2 x} + 16^{\cos^2 x} = 10 leads to sin⁡x=±12\sin x = \pm \frac{1}{2} and sin⁡x=±32\sin x = \pm \frac{\sqrt{3}}{2}, which have a total of 8 solutions in the interval [0,2π][0, 2\pi].
231 Medium
231 of 294
Application +4 | -1
If tan⁡(A+B)=3tan⁡A\tan ({A}+{B})=3 \tan {A}, then sin⁡2(A+B)+sin⁡2Asin⁡2B=\frac{\sin 2({A}+{B})+\sin 2 {A}}{\sin 2 {B}}=
View answer and explanation
Correct option: A
2
Option A (2) is correct because it represents the value of the expression sin⁡2(A+B)+sin⁡2Asin⁡2B\frac{{\sin 2(A+B)+\sin 2 A}}{{\sin 2 B}} given the condition tan⁡(A+B)=3tan⁡A\tan (A+B)=3 \tan A, which simplifies to 2 using trigonometric identities.
232 Medium
232 of 294
Application +4 | -1
If −π≤x≤π,−π≤y≤π-\pi \leq x \leq \pi,-\pi \leq y \leq \pi and cos⁡x+cos⁡y=2\cos x+\cos y=2, then the value of cos⁡(x−y)\cos (x-y) is
View answer and explanation
Correct option: C
1
Option C (1) is correct because the maximum value of cos⁡x\cos x is 1, so both cos⁡x\cos x and cos⁡y\cos y must be 1, leading to x=0x=0 and y=0y=0, and thus cos⁡(x−y)=1\cos(x-y) = 1.
233 Medium
233 of 294
Application +4 | -1
If m=cosec⁡θ−sin⁡θm=\operatorname{cosec} \theta-\sin \theta and n=sec⁡θ−cos⁡θn=\sec \theta-\cos \theta, then m2/3+n2/3=\mathrm{m}^{2 / 3}+\mathrm{n}^{2 / 3}=
View answer and explanation
Correct option: A
(mn)−2/3(\mathrm{mn})^{-2 / 3}
Option C ((mn)−2/3(mn)^{-2 / 3}) is correct because simplifying the expressions for m and n and then substituting them into the target expression yields the result.
234 Medium
234 of 294
Analyzing +4 | -1
The sum of the series sin⁡θsec⁡3θ+sin⁡3θsec⁡32θ+sin⁡32θsec⁡33θ\sin \theta \sec 3 \theta+\sin 3 \theta \sec 3^2 \theta+\sin 3^2 \theta \sec 3^3 \theta +…n+\ldots n terms, is
View answer and explanation
Correct option: C
12[tan⁡3nθ−tan⁡θ]\frac{1}{2}\left[\tan 3^n \theta-\tan \theta\right]
Option C, 12[tan⁡3nθ−tan⁡θ]\frac{1}{2}\left[\tan 3^n \theta-\tan \theta\right], is correct because the series telescopes, leaving only the first and last terms after simplification.
235 Medium
235 of 294
Application +4 | -1
The sum of all the 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: C
110π3\frac{110 \pi}{3}
None of the options are correct. The sum of all solutions is 30π30\pi, so Option D (None of these) is correct.
236 Medium
236 of 294
Application +4 | -1
If 2tan⁡2θ=sec⁡2θ2 \tan ^2 \theta=\sec ^2 \theta, then the general value of θ\theta is
View answer and explanation
Correct option: C
nπ±π4n \pi \pm \frac{\pi}{4}
Option C (nπ±π4n \pi \pm \frac{{\pi}}{{4}}) is correct because it represents the general solution derived from solving the given trigonometric equation.
237 Medium
237 of 294
Application +4 | -1
Let nn be a 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: D
4<n<84 < n < 8
Option D (4<n<84 < n < 8) is correct because by squaring the given equation and substituting n=6, the equation is satisfied.
238 Medium
238 of 294
Analyzing +4 | -1
The equation e∣sin⁡x∣+e−∣sin⁡x∣+4k=0e^{|\sin x|}+e^{-|\sin x|}+4 k=0 will have exactly four distinct solutions in [0,2π][0,2 \pi] if
View answer and explanation
Correct option: D
none of these
Option D (none of these) is correct because the given equation has no solution for k in the specified ranges.
239 Medium
239 of 294
Application +4 | -1
If f(θ)=12−3cos⁡θf(\theta)=\frac{1}{2-3 \cos \theta}, then range of f(θ)f(\theta) is
View answer and explanation
Correct option: C
(−∞,−1]∪(15,∞)(-\infty,-1] \cup\left(\frac{1}{5}, \infty\right)
Option D ((−∞,−1]∪(15,∞)( -\infty, -1] \cup (\frac{1}{5}, \infty)) is correct because by analyzing the range of 2−3cos⁡θ2-3\cos \theta, we find the range of f(θ)f(\theta) to be (−∞,−1]∪[15,∞)(-\infty, -1] \cup [\frac{1}{5}, \infty) excluding 0.
240 Medium
240 of 294
Analyzing +4 | -1
The number of values of θ\theta in the interval (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) satisfying the equation (1−tan⁡θ)(1+tan⁡θ)sec⁡2θ+2tan⁡2θ=0(1-\tan \theta)(1+\tan \theta) \sec ^2 \theta+2^{\tan ^2 \theta}=0 is
View answer and explanation
Correct option: C
4
Option D (None of these) is correct because the equation simplifies to a form where there are no solutions for θ in the given interval.
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