JEE Mathematics Trigonometric Equations MCQs Set 05

Answer: (c) \( \theta=\frac{4k+1}{n(n+1)}\cdot\frac{\pi}{2} \)

Explanation:
1. Write \( 2\cos r^2\theta\sin r\theta=\sin r(r+1)\theta-\sin r(r-1)\theta \).
2. The sum telescopes: \( \sum_{r=1}^{n}[\sin r(r+1)\theta-\sin r(r-1)\theta]=\sin n(n+1)\theta \), so \( \sin n(n+1)\theta=1 \).
3. \( n(n+1)\theta=2k\pi+\frac{\pi}{2}\Rightarrow\theta=\frac{4k+1}{n(n+1)}\cdot\frac{\pi}{2} \).

Teacher's Note:
1. Use 2 sin A cos B = sin(A + B) + sin(A − B) with A = rθ and B = r²θ.
2. Telescoping leaves only the last term.

Answer: (d) 9

Explanation:
1. Using product-to-sum, \( \sin8\theta+\sin2\theta=\sin16\theta+\sin2\theta \), so \( \sin16\theta=\sin8\theta \).
2. Either \( 16\theta=8\theta+2m\pi\Rightarrow\theta=\frac{m\pi}{4} \) or \( 16\theta=\pi-8\theta+2m\pi\Rightarrow\theta=\frac{(2m+1)\pi}{24} \).
3. In \( \left[0,\frac\pi2\right] \): \( 0,\frac\pi4,\frac\pi2 \) from the first family and \( \frac{\pi}{24},\frac{3\pi}{24},\frac{5\pi}{24},\frac{7\pi}{24},\frac{9\pi}{24},\frac{11\pi}{24} \) from the second. None coincide (6/24 is not an odd multiple of 1/24), so 3 + 6 = 9.

Teacher's Note:
1. Convert each product to a sum, cancel the common sin 2θ and solve sin A = sin B.
2. Check for repeated values between the two families.

Answer: (c) 2

Explanation:
1. Multiply by \( \cos x \) (with cos x ≠ 0): \( \sin x+1=2\cos^2x=2-2\sin^2x\Rightarrow2\sin^2x+\sin x-1=0 \).
2. \( (2\sin x-1)(\sin x+1)=0\Rightarrow\sin x=\frac12\ \text{or}\ -1 \).
3. \( \sin x=-1 \) gives \( x=\frac{3\pi}{2} \) where tan and sec are undefined, so it is rejected. \( \sin x=\frac12 \) gives \( x=\frac\pi6,\frac{5\pi}{6} \): 2 solutions.

Teacher's Note:
1. Check each root against the domain of tan x and sec x.
2. Cancelling cos x is fine here because cos x = 0 is not in the domain.

Answer: (c) \( (2n+1)\frac{\pi}{8},\ (2n+1)\frac{\pi}{4},\ (2n+1)\frac{\pi}{2} \)

Explanation:
1. Group: \( (\cos7\theta+\cos\theta)+(\cos5\theta+\cos3\theta)=2\cos4\theta\cos3\theta+2\cos4\theta\cos\theta \).
2. \( =2\cos4\theta(\cos3\theta+\cos\theta)=4\cos4\theta\cos2\theta\cos\theta=0 \).
3. \( \cos4\theta=0,\ \cos2\theta=0,\ \cos\theta=0 \) give \( (2n+1)\frac\pi8,\ (2n+1)\frac\pi4,\ (2n+1)\frac\pi2 \).

Teacher's Note:
1. Pair terms with equal angle sums so that a common cos 4θ appears.
2. Each factor gives odd multiples of a different angle.

Answer: (a) 2

Explanation:
1. Multiply by \( \cos\theta \): \( 5-13\cos\theta=12\sin\theta\Rightarrow13\cos\theta+12\sin\theta=5 \).
2. \( \sqrt{13^2+12^2}=\sqrt{313}\gt 5 \), so solutions exist: \( \cos(\theta-\alpha)=\frac{5}{\sqrt{313}} \).
3. Since the right side is between −1 and 1 (and not ±1), there are exactly two values of \( \theta-\alpha \) in each period, hence 2 solutions in \( [0,2\pi] \) (cos θ = 0 does not satisfy the equation).

Teacher's Note:
1. Reduce to a cos θ + b sin θ = c and compare |c| with √(a² + b²).
2. A value strictly between −1 and 1 is taken twice per period by the cosine.

Answer: (c) \( \frac{\pi}{6} \)

Explanation:
1. \( \sin\theta+1=\sqrt3\cos\theta\Rightarrow\sqrt3\cos\theta-\sin\theta=1\Rightarrow\cos\left(\theta+\frac\pi6\right)=\frac12 \).
2. \( \theta+\frac{\pi}{6}=2n\pi\pm\frac\pi3 \), so \( \theta=2n\pi+\frac\pi6 \) or \( \theta=2n\pi-\frac\pi2 \).
3. In \( (0,2\pi] \): \( \frac\pi6 \) and \( \frac{3\pi}{2} \). At \( \frac{3\pi}{2} \) tan and sec are undefined, so the only value is \( \frac\pi6 \).

Teacher's Note:
1. After multiplying by cos θ, always re-check roots where cos θ = 0.
2. Write a cos θ + b sin θ as a single cosine.

Answer: (c) \( \left[\frac{3\pi}{2},2\pi\right]\cup\{0\} \)

Explanation:
1. \( \cos x-\sin x=\sqrt2\cos\left(x+\frac\pi4\right)\ge1\Rightarrow\cos\left(x+\frac\pi4\right)\ge\frac{1}{\sqrt2} \).
2. \( -\frac\pi4\le x+\frac\pi4\le\frac\pi4 \) (plus multiples of 2π), so \( -\frac\pi2\le x\le0 \) modulo 2π.
3. In \( [0,2\pi] \): \( x=0 \) and \( \frac{3\pi}{2}\le x\le2\pi \).

Teacher's Note:
1. Write cos x − sin x as √2 cos(x + π/4).
2. Shift the interval by 2π to bring it into [0, 2π].

Answer: (c) \( (2n+1)\frac\pi2 \)

Explanation:
1. The left side is at most 1 and the right side is at least 1, so both equal 1.
2. \( \sin^6x=1\Rightarrow\cos x=0\Rightarrow x=(2n+1)\frac\pi2 \) and \( \cos^43x=0 \).
3. At \( x=(2n+1)\frac\pi2 \), \( \cos3x=4\cos^3x-3\cos x=0 \) as well, so all such x work.

Teacher's Note:
1. Compare the maximum of one side with the minimum of the other.
2. Verify the second condition with cos 3x = 4cos³x − 3cos x.

Answer: (a) \( \varnothing \) (empty set)

Explanation:
1. The left side is \( \tan(3x-2x)=\tan x \), so \( \tan x=1\Rightarrow x=n\pi+\frac\pi4 \).
2. For these values \( 2x=2n\pi+\frac\pi2 \), so \( \tan2x \) is not defined and the original expression does not exist.
3. Therefore no value of x satisfies the equation.

Teacher's Note:
1. The formula tan(A − B) holds only when tan A, tan B and the denominator are defined.
2. Always check the domain of the original expression.

Answer: (a) \( \frac\pi4 \)

Explanation:
1. \( \sin x\cdot2\sqrt2|\cos x|=1\Rightarrow\sin x|\cos x|=\frac{1}{2\sqrt2} \), which needs \( \sin x\gt 0 \).
2. If \( \cos x\gt 0 \): \( \sin2x=\frac{1}{\sqrt2}\Rightarrow x=\frac\pi8,\frac{3\pi}{8} \). If \( \cos x\lt 0 \): \( \sin2x=-\frac1{\sqrt2}\Rightarrow x=\frac{5\pi}{8},\frac{7\pi}{8} \).
3. The values \( \frac\pi8,\frac{3\pi}8,\frac{5\pi}8,\frac{7\pi}8 \) have common difference \( \frac\pi4 \).

Teacher's Note:
1. Treat the cases cos x > 0 and cos x < 0 separately because of the modulus.
2. Note sin x must be positive for the product to be positive.

Answer: (a) \( \theta=n\pi\ \text{or}\ n\pi+(-1)^n\frac\pi6 \)

Explanation:
1. A homogeneous system has a non-trivial solution iff the determinant is zero: \( \begin{vmatrix}\sin3\theta&-1&1\\\cos2\theta&4&3\\2&7&7\end{vmatrix}=7\sin3\theta+14\cos2\theta-14=0 \).
2. \( \sin3\theta+2\cos2\theta-2=0\Rightarrow\sin3\theta=4\sin^2\theta\Rightarrow\sin\theta(4\sin^2\theta+4\sin\theta-3)=0 \).
3. \( (2\sin\theta-1)(2\sin\theta+3)=0 \): \( \sin\theta=0 \) or \( \sin\theta=\frac12 \), so \( \theta=n\pi \) or \( n\pi+(-1)^n\frac\pi6 \).

Teacher's Note:
1. Set the determinant of the coefficients to zero.
2. Use sin 3θ = 3 sin θ − 4 sin³θ and cos 2θ = 1 − 2 sin²θ.

Answer: (c) \( a\in(-\infty,-2] \)

Explanation:
1. Let \( t=\cos^2x\in[0,1] \). Then \( t^2+at+1=0 \), so \( a=-\left(t+\frac1t\right) \) for \( t\in(0,1] \).
2. \( t+\frac1t\ge2 \) for positive t, so \( a\le-2 \).
3. Every value \( a\le-2 \) is attained for some \( t\in(0,1] \), so \( a\in(-\infty,-2] \).

Teacher's Note:
1. Isolate a and use the AM-GM inequality on t + 1/t.
2. The restriction t ≤ 1 still allows all values of t + 1/t from 2 upwards.

Answer: (b) 1

Explanation:
1. Put \( x=0 \) and \( x=\pi \): \( a_1+a_3+a_5=0 \) and \( a_1-a_3+a_5=0 \), so \( a_3=0,\ a_1+a_5=0 \).
2. Put \( x=\frac\pi2 \) and \( x=\frac{3\pi}2 \): \( a_1+a_2-a_5=0 \) and \( a_1-a_2-a_5=0 \), so \( a_2=0,\ a_1=a_5 \).
3. Hence \( a_1=a_5=0 \). Then \( a_4\sin2x=0 \) for all x gives \( a_4=0 \). Only the zero 5-tuple works: 1.

Teacher's Note:
1. Substitute convenient values of x to build a linear system for the coefficients.
2. The functions 1, sin x, cos x, sin 2x, cos 2x are linearly independent.

Answer: (a) \( x=n\pi+(-1)^n\frac\pi6,\ y=m\pi\pm\frac\pi6 \)

Explanation:
1. First equation: \( \csc^2x=3\sec^2y \). Second: \( 2\csc x+\sqrt3|\sec y|=6 \).
2. Let \( a=\csc x,\ b=|\sec y| \) with \( a^2=3b^2 \). If \( a=\sqrt3b \): \( 2\sqrt3b+\sqrt3b=6\Rightarrow b=\frac{2}{\sqrt3},\ a=2 \). If \( a=-\sqrt3b \) then \( -\sqrt3b=6 \), impossible.
3. \( \csc x=2\Rightarrow\sin x=\frac12\Rightarrow x=n\pi+(-1)^n\frac\pi6 \); \( |\sec y|=\frac2{\sqrt3}\Rightarrow\cos y=\pm\frac{\sqrt3}{2}\Rightarrow y=m\pi\pm\frac\pi6 \).

Teacher's Note:
1. Equal powers of the same base must have equal exponents, so 5^0 = 1 gives exponent 0 and 64 = 2^6 gives exponent 6.
2. Solve the resulting algebraic system in csc x and |sec y|.

Answer: (c) two

Explanation:
1. First: \( 2\sin^2\theta-(1-2\sin^2\theta)=0\Rightarrow\sin^2\theta=\frac14\Rightarrow\sin\theta=\pm\frac12 \).
2. Second: \( 2-2\sin^2\theta-3\sin\theta=0\Rightarrow2\sin^2\theta+3\sin\theta-2=0\Rightarrow(2\sin\theta-1)(\sin\theta+2)=0 \), so \( \sin\theta=\frac12 \).
3. Common: \( \sin\theta=\frac12\Rightarrow\theta=\frac\pi6,\frac{5\pi}6 \): 2 solutions.

Teacher's Note:
1. Solve each equation separately in terms of sin θ and keep only the common values.
2. sin θ = −2 is impossible.

Answer: (c) 0

Explanation:
1. Multiply by \( \sin x\cos x \): \( \sqrt3\sin x+\cos x+2(\sin^2x-\cos^2x)=0\Rightarrow2\sin\left(x+\frac\pi6\right)=2\cos2x \).
2. \( \sin\left(x+\frac\pi6\right)=\sin\left(\frac\pi2-2x\right) \): either \( x=\frac\pi9+\frac{2k\pi}3 \) or \( x=-\frac\pi3+2k\pi \).
3. In \( (-\pi,\pi) \): \( \frac\pi9,\frac{7\pi}9,-\frac{5\pi}9,-\frac\pi3 \). Sum \( =\frac{\pi+7\pi-5\pi-3\pi}{9}=0 \).

Teacher's Note:
1. Clear the denominators by multiplying by sin x cos x.
2. Use sin A = sin B for both families of solutions.

Answer: (c) \( \frac{13}{9} \)

Explanation:
1. \( \cos\left(\frac\pi6+x\right)\cos\left(\frac\pi6-x\right)=\frac12\left(\cos\frac\pi3+\cos2x\right)=\frac14+\frac12\cos2x \).
2. The equation becomes \( 8\cos x\left(\frac12\cos2x-\frac14\right)=1\Rightarrow4\cos x\cos2x-2\cos x=1\Rightarrow2\cos3x=1 \).
3. \( \cos3x=\frac12\Rightarrow3x=\frac\pi3,\frac{5\pi}3,\frac{7\pi}3\Rightarrow x=\frac\pi9,\frac{5\pi}9,\frac{7\pi}9 \). Sum \( =\frac{13\pi}9 \), so \( k=\frac{13}{9} \).

Teacher's Note:
1. Use the product formula 2 cos A cos B = cos(A + B) + cos(A − B).
2. Then 2 cos x cos 2x = cos 3x + cos x.

Answer: (a) 5

Explanation:
1. The left side is at least 1 and the right side is at most 1, so both equal 1.
2. \( \sin x=0\Rightarrow x=n\pi \), and then \( \cos^23x=\cos^23n\pi=1 \) holds.
3. In \( \left[-\frac{5\pi}2,\frac{5\pi}2\right] \): \( x=-2\pi,-\pi,0,\pi,2\pi \): 5 solutions.

Teacher's Note:
1. Use the bounds of both sides.
2. Count multiples of π inside the interval.

Answer: (c) (Q) and (T)

Explanation:
1. \( f'(x)=-\pi\sin x\cos(\pi\cos x)=0 \): either \( \sin x=0 \) or \( \cos(\pi\cos x)=0\Rightarrow\cos x=\pm\frac12 \).
2. So \( Y \) consists of all positive multiples of \( \frac\pi3 \): \( \frac\pi3,\frac{2\pi}3,\pi,\frac{4\pi}3,\dots \).
3. This is an arithmetic progression (Q) and contains \( \frac\pi3,\frac{2\pi}3,\pi \) (T). It does not contain \( \frac\pi2 \) (P) or \( \frac\pi6 \) (U).

Teacher's Note:
1. Find where each derivative vanishes and list the first few elements.
2. Test every property against that list.

Answer: (b) (P), (R) and (S)

Explanation:
1. \( g'(x)=-2\pi\cos x\sin(2\pi\sin x)=0 \): either \( \cos x=0 \) or \( 2\pi\sin x=k\pi \), i.e. \( \sin x\in\{0,\pm\frac12,\pm1\} \).
2. So \( W \) contains \( \frac\pi2,\frac{3\pi}2 \) (cos x = 0), \( 4\pi,7\pi \) (sin x = 0) so (P) holds, and \( \frac\pi6,\frac{7\pi}6,\frac{13\pi}6 \) (sin x = ±1/2) so (S) holds.
3. The first elements \( \frac\pi6,\frac\pi2,\frac{5\pi}6,\pi,\dots \) have unequal gaps, so W is not an A.P. (R). \( \sin\frac\pi3=\frac{\sqrt3}2 \) is not allowed, so (T) fails.

Teacher's Note:
1. Set g'(x) = 0 and note that sin(2π sin x) = 0 needs 2 sin x to be an integer.
2. Test each property on the list of elements.

Answer: (b) \( \left[-1,-\frac12\right] \)

Explanation:
1. \( \cos^4\theta+\sin^4\theta=1-\frac12\sin^22\theta \).
2. Since \( 0\le\sin^22\theta\le1 \), the value lies in \( \left[\frac12,1\right] \).
3. \( \lambda=-\left(1-\frac12\sin^22\theta\right)\in\left[-1,-\frac12\right] \).

Teacher's Note:
1. Write cos⁴θ + sin⁴θ = (cos²θ + sin²θ)² − 2 sin²θ cos²θ.
2. Use the range of sin² 2θ.

Answer: (b) 1

Explanation:
1. Let \( \theta=\pi x-\frac\pi4 \). Then \( 3\pi x+\frac\pi4=3\theta+\pi \), so \( \sin\left(3\pi x+\frac\pi4\right)=-\sin3\theta \).
2. \( f=(3-\sin2\pi x)\sin\theta+\sin3\theta=\sin\theta\,(6-\sin2\pi x-4\sin^2\theta) \). Since \( 4\sin^2\theta=2-2\cos2\theta=2-2\sin2\pi x \), the bracket equals \( 4+\sin2\pi x\gt 0 \).
3. So \( f(x)\ge0\iff\sin\left(\pi x-\frac\pi4\right)\ge0 \). For \( x\in[0,2] \) this means \( 0\le\pi x-\frac\pi4\le\pi \), i.e. \( \frac14\le x\le\frac54 \), so \( \beta-\alpha=1 \).

Teacher's Note:
1. Rewrite the second sine as a triple-angle expression of the first angle and factor.
2. The remaining factor is always positive, so only the sign of sin θ matters.