JEE Mathematics Trigonometric Equations MCQs Set 03

Answer: (b) \( x=2n\pi+\frac{\pi}{6} \)

Explanation:
1. The logarithms exist only if \( \sin x>0 \) and \( \cos x>0 \), so x lies in the first quadrant.
2. Using \( a^{\log_a t}=t \): \( \sqrt5+\sqrt5\sin x=\sqrt{15}\cos x \), i.e. \( 1+\sin x=\sqrt3\cos x \).
3. Rewrite as \( \frac{\sqrt3}{2}\cos x-\frac12\sin x=\frac12 \), so \( \cos\left(x+\frac{\pi}{6}\right)=\cos\frac{\pi}{3} \).
4. Hence \( x=2n\pi+\frac{\pi}{6} \) or \( x=2n\pi-\frac{\pi}{2} \); the second has \( \sin x<0 \) and is rejected.

Teacher's Note:
1. Always apply the domain conditions of logarithms before accepting roots.
2. The first-quadrant restriction removes the root coming from the minus sign.
3. Check the answer by substituting x = 30 degrees: both sides equal about 3.354.

Answer: (d) 6

Explanation:
1. \( \sin5\theta-\sin3\theta=2\cos4\theta\sin\theta \), so the equation becomes \( \sin\theta(1+2\cos4\theta)=0 \).
2. \( \sin\theta=0 \) gives \( \theta=0,\pi \).
3. \( \cos4\theta=-\frac12 \) gives \( 4\theta=\frac{2\pi}{3},\frac{4\pi}{3},\frac{8\pi}{3},\frac{10\pi}{3} \), i.e. \( \theta=\frac{\pi}{6},\frac{\pi}{3},\frac{2\pi}{3},\frac{5\pi}{6} \).
4. Total number of solutions = 2 + 4 = 6.

Teacher's Note:
1. Reduce sums of sines to products to get a factorised equation.
2. Do not forget the endpoints 0 and pi when the interval is closed.

Answer: (a) \( x=\frac{n\pi}{7}-\frac{\pi}{84} \)

Explanation:
1. Divide \( \cos3x+\sqrt3\sin3x \) by 2: it equals \( 2\sin\left(3x+\frac{\pi}{6}\right) \), so the equation is \( \sin11x+\sin\left(3x+\frac{\pi}{6}\right)=0 \).
2. Sum to product: \( 2\sin\left(7x+\frac{\pi}{12}\right)\cos\left(4x-\frac{\pi}{12}\right)=0 \).
3. \( \sin\left(7x+\frac{\pi}{12}\right)=0\Rightarrow x=\frac{n\pi}{7}-\frac{\pi}{84} \).
4. \( \cos\left(4x-\frac{\pi}{12}\right)=0\Rightarrow x=\frac{n\pi}{4}+\frac{7\pi}{48} \) (the other family).

Teacher's Note:
1. Write a cos + b sin as a single sine first.
2. Two factors give two families of solutions; any one family is a valid answer here.

Answer: (c) 6

Explanation:
1. \( \cos\theta\cos3\theta=\frac12(\cos4\theta+\cos2\theta) \), so \( (\cos4\theta+\cos2\theta)\cos2\theta=\frac12 \).
2. Expand: \( 2\cos^22\theta+2\cos2\theta\cos4\theta-1=0 \), and \( 2\cos^22\theta-1=\cos4\theta \), giving \( \cos4\theta(1+2\cos2\theta)=0 \).
3. \( \cos4\theta=0\Rightarrow\theta=\frac{\pi}{8},\frac{3\pi}{8},\frac{5\pi}{8},\frac{7\pi}{8} \).
4. \( \cos2\theta=-\frac12\Rightarrow\theta=\frac{\pi}{3},\frac{2\pi}{3} \). Total 6.

Teacher's Note:
1. Convert products into sums stepwise.
2. Count the roots of each factor separately within the given interval.

Answer: (c) \( x=2n\pi\pm\frac{2\pi}{3} \)

Explanation:
1. Put \( y=\tan x\cot\frac x2 \), so the equation is \( 2+y+\frac1y=0 \), i.e. \( (y+1)^2=0 \) and \( y=-1 \).
2. \( \tan x\cot\frac x2=\frac{2\sin\frac x2\cos\frac x2}{\cos x}\cdot\frac{\cos\frac x2}{\sin\frac x2}=\frac{2\cos^2\frac x2}{\cos x}=\frac{1+\cos x}{\cos x} \).
3. \( \frac{1+\cos x}{\cos x}=-1\Rightarrow\cos x=-\frac12 \).
4. So \( x=2n\pi\pm\frac{2\pi}{3} \).

Teacher's Note:
1. A substitution y + 1/y = -2 forces y = -1.
2. Check that tan(x/2), cot(x/2) are defined at the answer: they are.

Answer: (a) 4

Explanation:
1. Squaring (with \( 6\tan x-3\ge0 \)): \( 13-18\tan x=36\tan^2x-36\tan x+9 \), so \( 36\tan^2x-18\tan x-4=0 \).
2. \( (3\tan x-2)(6\tan x+1)=0 \) gives \( \tan x=\frac23 \) or \( \tan x=-\frac16 \).
3. \( \tan x=-\frac16 \) makes \( 6\tan x-3=-4<0 \), so it is rejected.
4. \( \tan x=\frac23 \) gives \( x=\alpha,\ \pi+\alpha,\ \alpha-\pi,\ \alpha-2\pi \) in the interval: 4 solutions.

Teacher's Note:
1. After squaring, test the root against the sign of the right-hand side.
2. Each value of tan x occurs twice per period of 2 pi, so there are four in (-2 pi, 2 pi).

Answer: (d) \( \frac{\pi}{16} \)

Explanation:
1. Rearrange: \( \sqrt{1+\sin2x}=\sqrt2\cos3x \), so \( 1+\sin2x=2\cos^23x=1+\cos6x \) with \( \cos3x\ge0 \).
2. \( \sin2x=\cos6x\Rightarrow\cos\left(\frac{\pi}{2}-2x\right)=\cos6x \).
3. \( \frac{\pi}{2}-2x=2n\pi\pm6x \) gives \( x=\frac{n\pi}{4}+\frac{\pi}{16} \) or \( x=\frac{n\pi}{2}-\frac{\pi}{4} \).
4. Smallest positive value: \( x=\frac{\pi}{16} \), and \( \cos\frac{3\pi}{16}>0 \) so it is valid.

Teacher's Note:
1. Keep the condition cos 3x >= 0 in mind after squaring.
2. Try n = 0 first when looking for the least positive root.

Answer: (b) \( \frac{\pi}{12} \)

Explanation:
1. Write \( 2\sin\left(3x+\frac{\pi}{4}\right)=\sqrt2(\sin3x+\cos3x) \) and square: \( 2(1+\sin6x)=1+8\sin2x\cos^22x \).
2. \( 8\sin2x\cos^22x=4\sin2x(1+\cos4x)=4\sin2x+2(\sin6x-\sin2x) \), so the right side is \( 1+2\sin2x+2\sin6x \).
3. Equating: \( 2+2\sin6x=1+2\sin2x+2\sin6x \), so \( \sin2x=\frac12 \) and \( x=\frac{\pi}{12},\frac{5\pi}{12},\frac{13\pi}{12},\dots \).
4. The left side must be non-negative. At \( x=\frac{\pi}{12} \): \( 2\sin\frac{\pi}{2}=2>0 \) (valid). At \( x=\frac{5\pi}{12} \): \( 2\sin\frac{3\pi}{2}<0 \) (rejected). So the smallest is \( \frac{\pi}{12} \).

Teacher's Note:
1. Squaring creates extraneous roots; always test the sign of the left side.
2. Here pi/12 is valid but 5 pi/12 is not.

Answer: (c) 10

Explanation:
1. Group: \( (\sin x+\sin5x)-\sin3x=2\sin3x\cos2x-\sin3x=\sin3x(2\cos2x-1) \); similarly the right side is \( \cos3x(2\cos2x-1) \).
2. So \( (2\cos2x-1)(\sin3x-\cos3x)=0 \).
3. \( \cos2x=\frac12 \) gives \( x=\frac{\pi}{6},\frac{5\pi}{6},\frac{7\pi}{6},\frac{11\pi}{6} \) (4 values).
4. \( \tan3x=1 \) gives \( x=\frac{\pi}{12}+\frac{n\pi}{3} \) (6 values in \( [0,2\pi) \)). Total 10.

Teacher's Note:
1. The two groups of solutions do not overlap, so add their counts.
2. Principal solutions are those in [0, 2 pi).

Answer: (a) \( x=2n\pi+\frac{\pi}{12} \)

Explanation:
1. Domain: \( \cos x+\sin x>0 \) and \( \cos x-\sin x>0 \), which means \( -\frac{\pi}{4}<x<\frac{\pi}{4} \) (mod \( 2\pi \)).
2. The equation becomes \( \sqrt3(\cos x+\sin x)-(\cos x-\sin x)=\sqrt2 \), i.e. \( (\sqrt3-1)\cos x+(\sqrt3+1)\sin x=\sqrt2 \).
3. Divide by \( 2\sqrt2 \): \( \cos(x-75^\circ)=\frac12 \), so \( x=75^\circ\pm60^\circ+2n\pi \), i.e. \( x=2n\pi+\frac{\pi}{12} \) or \( x=2n\pi+\frac{3\pi}{4} \).
4. \( x=\frac{3\pi}{4} \) lies outside the domain, so only \( x=2n\pi+\frac{\pi}{12} \) is valid.

Teacher's Note:
1. Logarithmic equations need the domain check at the end.
2. Here cos x - sin x is negative at 3 pi/4, so that root must be discarded.

Answer: (b) 5

Explanation:
1. \( \cos6\theta+\cos2\theta=2\cos4\theta\cos2\theta \), so the equation is \( \cos4\theta(2\cos2\theta+1)+1=0 \).
2. With \( c=\cos2\theta \): \( (2c^2-1)(2c+1)+1=4c^3+2c^2-2c=2c(2c-1)(c+1)=0 \).
3. \( \cos2\theta=0\Rightarrow\theta=45^\circ,135^\circ \); \( \cos2\theta=\frac12\Rightarrow\theta=30^\circ,150^\circ \); \( \cos2\theta=-1\Rightarrow\theta=90^\circ \).
4. Total 5 solutions.

Teacher's Note:
1. Substituting c = cos 2 theta turns the equation into a cubic.
2. Remember that every root of cos 2 theta must be converted back into theta values in the interval.

Answer: (d) \( \left\{-\frac{5\pi}{3}\right\} \)

Explanation:
1. The base must be positive and not 1: \( 0<-x^2-6x \) gives \( -6<x<0 \) (and the base is never 1 here).
2. Equal logs need equal arguments, both positive: \( \sin3x+\sin x=2\sin2x\cos x=\sin2x \) with \( \sin2x>0 \).
3. So \( \cos x=\frac12 \), which gives \( x=-\frac{5\pi}{3},-\frac{\pi}{3} \) in \( (-6,0) \).
4. \( \sin2x>0 \) at \( x=-\frac{5\pi}{3} \) (it equals \( \frac{\sqrt3}{2} \)) but not at \( -\frac{\pi}{3} \) (it equals \( -\frac{\sqrt3}{2} \)). Only \( -\frac{5\pi}{3} \) works.

Teacher's Note:
1. Check base and argument conditions of every logarithm.
2. Here sin 2x must be positive, which removes -pi/3.

Answer: (b) \( \theta=2n\pi\ \text{or}\ \theta=2n\pi+\frac{\pi}{2} \)

Explanation:
1. Use \( \cos2\theta=1-2\sin^2\theta \): \( 2+2\sin^2\theta-2\cos\theta-4\sin\theta+2\sin\theta\cos\theta=0 \).
2. This is \( 2(\sin\theta-1)^2+2\cos\theta(\sin\theta-1)=0 \), so \( (\sin\theta-1)(\sin\theta-1+\cos\theta)=0 \).
3. \( \sin\theta=1\Rightarrow\theta=2n\pi+\frac{\pi}{2} \).
4. \( \sin\theta+\cos\theta=1\Rightarrow\sin\left(\theta+\frac{\pi}{4}\right)=\frac{1}{\sqrt2} \), so \( \theta=2n\pi \) or \( 2n\pi+\frac{\pi}{2} \). Together: \( \theta=2n\pi \) or \( 2n\pi+\frac{\pi}{2} \).

Teacher's Note:
1. Factor out (sin theta - 1) after using the double-angle identities.
2. Solve sin + cos = 1 by writing it as root 2 sin(theta + pi/4).

Answer: (d) 5025

Explanation:
1. \( \sin\pi x+\cos\pi x=\sqrt2\sin\left(\pi x+\frac{\pi}{4}\right)=0 \), so \( \pi x+\frac{\pi}{4}=n\pi \) and \( x=n-\frac14 \).
2. In \( [0,100] \) the values are \( n=1,2,\dots,100 \) (n = 0 gives a negative x).
3. Sum = \( \sum_{n=1}^{100}\left(n-\frac14\right)=5050-25=5025 \).

Teacher's Note:
1. List the first and last solutions to fix the range of n.
2. Use the formula for the sum of the first n natural numbers.

Answer: (a) 72

Explanation:
1. If \( a^3+b^3+c^3=(a+b+c)^3 \) then \( (a+b)(b+c)(c+a)=0 \).
2. \( \sin x+\sin2x=2\sin\frac{3x}{2}\cos\frac x2 \), \( \sin2x+\sin3x=2\sin\frac{5x}{2}\cos\frac x2 \), \( \sin3x+\sin x=2\sin2x\cos x \).
3. Positive roots: \( x=\frac{2\pi}{3} \), \( x=\frac{2\pi}{5} \), \( x=\pi \), \( x=\frac{\pi}{2} \), ...
4. The least is \( \frac{2\pi}{5}=72^\circ \).

Teacher's Note:
1. Use the identity (a + b + c)^3 - a^3 - b^3 - c^3 = 3(a + b)(b + c)(c + a).
2. Compare the smallest positive root of each factor.

Answer: (c) \( [-\sqrt2,\sqrt2] \)

Explanation:
1. \( y=\sin x-\cos x=\sqrt2\sin\left(x-\frac{\pi}{4}\right) \).
2. The sine takes all values in \( [-1,1] \), so \( y \) takes all values in \( [-\sqrt2,\sqrt2] \).

Teacher's Note:
1. Write a sin x + b cos x as R sin(x + phase), range [-R, R].
2. R = sqrt(a^2 + b^2).

Answer: (c) \( x=\frac{\pi}{2},\ \pi \)

Explanation:
1. The equation is \( \sin x-\cos x=1 \), i.e. \( \sqrt2\sin\left(x-\frac{\pi}{4}\right)=1 \).
2. \( x-\frac{\pi}{4}=\frac{\pi}{4} \) or \( \frac{3\pi}{4} \) (mod \( 2\pi \)), so \( x=\frac{\pi}{2} \) or \( x=\pi \).
3. Check: at \( x=\pi \), \( \sin\pi-\cos\pi=1 \). At \( \frac{3\pi}{2} \) the value is \( -1 \).

Teacher's Note:
1. Substitute candidate values back to be sure.
2. The interval is open, so 0 and 2 pi are excluded.

Answer: (a) \( \theta=2n\pi+\frac{\pi}{4}\ \text{or}\ \theta=2n\pi+\pi-\tan^{-1}2 \)

Explanation:
1. A quadratic is a perfect square when its discriminant is zero and its leading coefficient is positive: \( \sin\theta>0 \).
2. \( D=4\cos^2\theta-2\sin\theta(\cos\theta+\sin\theta)=0 \), i.e. \( 2\cos^2\theta-\sin\theta\cos\theta-\sin^2\theta=0 \).
3. Divide by \( \cos^2\theta \): \( \tan^2\theta+\tan\theta-2=0 \), so \( \tan\theta=1 \) or \( -2 \).
4. \( \tan\theta=1 \) with \( \sin\theta>0 \) gives \( \theta=2n\pi+\frac{\pi}{4} \); \( \tan\theta=-2 \) with \( \sin\theta>0 \) gives \( \theta=2n\pi+\pi-\tan^{-1}2 \).

Teacher's Note:
1. A negative leading coefficient would give minus a square, not a square.
2. The condition sin theta > 0 halves each family of roots.

Answer: (d) 0

Explanation:
1. Each factor lies in \( [-1,1] \), so the product equals 1 only if all three have absolute value 1 and an even number are \( -1 \).
2. If \( \sin x=\pm1 \) then \( \cos x=0 \) and \( \sin2x=2\sin x\cos x=0 \), contradicting \( |\sin2x|=1 \).
3. So no real x exists; the number of solutions is 0.

Teacher's Note:
1. Bound each factor by 1 and check whether the extreme values can occur together.
2. sin x = +/-1 and sin 2x = +/-1 cannot hold at the same x.

Answer: (b) \( -\frac32 \)

Explanation:
1. Let \( s=\sin^2x\cos^2x \). Then \( \sin^6x+\cos^6x=1-3s \) and \( \sin^4x+\cos^4x=1-2s \).
2. \( f(x)=1-3s+k(1-2s)=(1+k)-(3+2k)s \).
3. For \( f \) to be constant, the coefficient of \( s \) must vanish: \( 3+2k=0\Rightarrow k=-\frac32 \).
4. Then \( f(x)=1-\frac32=-\frac12 \).

Teacher's Note:
1. Express everything in terms of sin^2 x cos^2 x.
2. The value of the constant is also worth remembering: -1/2.

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

Explanation:
1. From \( f=0 \): \( k=\frac{3s-1}{1-2s} \) where \( s=\sin^2c\cos^2c=\frac14\sin^22c\in\left[0,\frac14\right] \).
2. The function \( g(s)=\frac{3s-1}{1-2s} \) is increasing on this interval (its derivative is \( \frac{1}{(1-2s)^2}>0 \)).
3. \( g(0)=-1 \) and \( g\left(\frac14\right)=\frac{-1/4}{1/2}=-\frac12 \).
4. Hence \( k\in\left[-1,-\frac12\right] \).

Teacher's Note:
1. Find the range of s first, then the range of k as a function of s.
2. Check monotonicity before substituting end values.

Answer: (a) \( \theta=\frac{n\pi}{2}\pm\frac{\pi}{6} \)

Explanation:
1. With \( k=-0.7 \): \( 1-3s-0.7(1-2s)=0 \), so \( 0.3-1.6s=0 \) and \( s=\frac{3}{16} \).
2. \( s=\frac14\sin^22\theta \) gives \( \sin^22\theta=\frac34 \).
3. \( \sin^22\theta=\sin^2\frac{\pi}{3}\Rightarrow2\theta=n\pi\pm\frac{\pi}{3} \).
4. Therefore \( \theta=\frac{n\pi}{2}\pm\frac{\pi}{6} \).

Teacher's Note:
1. Use sin^2 A = sin^2 B => A = n pi +/- B to keep both signs of sin 2 theta.
2. Taking only sin 2 theta = +root3/2 would lose half the solutions.

Answer: (b) A-R, B-S, C-P, D-Q

Explanation:
1. Write \( a\cos\theta=c-b\sin\theta \), square it: \( (a^2+b^2)\sin^2\theta-2bc\sin\theta+(c^2-a^2)=0 \).
2. Roots \( \sin\alpha,\sin\beta \): sum \( \frac{2bc}{a^2+b^2} \) (A-R) and product \( \frac{c^2-a^2}{a^2+b^2} \) (B-S).
3. Use \( t=\tan\frac\theta2 \): \( a\frac{1-t^2}{1+t^2}+b\frac{2t}{1+t^2}=c \) gives \( (a+c)t^2-2bt+(c-a)=0 \).
4. Sum of roots \( \frac{2b}{a+c} \) (C-P) and product \( \frac{c-a}{c+a} \) (D-Q).

Teacher's Note:
1. Squaring isolates a quadratic in sin theta; the half-angle substitution gives one in tan(theta/2).
2. Sum and product of roots give the required expressions.

Answer: (d) A-S, B-P, C-Q, D-R

Explanation:
1. (A) \( \cos3x\cos^3x+\sin3x\sin^3x=\cos^32x \), so \( \cos2x=0 \) and \( x=\frac{n\pi}{2}\pm\frac\pi4 \) (S).
2. (B) \( 4\sin\alpha\sin(x+\alpha)\sin(x-\alpha)=2\sin\alpha(\cos2\alpha-\cos2x) \) and \( \sin3\alpha=3\sin\alpha-4\sin^3\alpha=2\sin\alpha(\cos2\alpha+\tfrac12) \), so \( \cos2x=-\frac12 \) and \( x=n\pi\pm\frac\pi3 \) (P).
3. (C) With \( t=\tan x \): \( |2t-1|+\left|\frac2t-1\right|\ge2 \) with equality only at \( t=1 \) (checked numerically), so \( x=n\pi+\frac\pi4 \) (Q).
4. (D) Put \( c=\cos2x \): \( \left(\frac{1-c}{2}\right)^5+\left(\frac{1+c}{2}\right)^5=\frac{1+10c^2+5c^4}{16}=\frac{29}{16}c^4 \) gives \( 24c^4-10c^2-1=0 \), so \( c^2=\frac12 \) (the other root \( c^2=-\frac1{12} \) is impossible). Then \( \cos4x=2c^2-1=0 \), so \( x=\frac{n\pi}{4}+\frac\pi8 \) (R).

Teacher's Note:
1. Identify each equation with a standard identity before solving.
2. For part C, both absolute values are non-negative, and the sum 2 is attained only when tan x = 1.

Answer: (b) \( x=\left(\frac{11\pi}{6}\right)^2 \)

Explanation:
1. Let \( a=\frac{\sqrt x}{2} \): \( \sin a+\cos a=\sqrt2\sin\left(a+\frac\pi4\right) \), so the equation is \( \sin\left(a+\frac\pi4\right)=\sin2a \).
2. Case 1: \( 2a=a+\frac\pi4+2n\pi\Rightarrow a=\frac\pi4+2n\pi \), so \( \sqrt x=\frac\pi2+4n\pi \).
3. Case 2: \( 2a=\pi-a-\frac\pi4+2n\pi\Rightarrow a=\frac\pi4+\frac{2n\pi}{3} \), so \( \sqrt x=\frac\pi2+\frac{4n\pi}{3} \) (this contains Case 1).
4. For \( n=0,1,2,\dots \) the values of \( \sqrt x \) are \( \frac\pi2,\ \frac{11\pi}{6},\ \frac{19\pi}{6},\dots \), so the second solution is \( x=\left(\frac{11\pi}{6}\right)^2 \).

Teacher's Note:
1. Avoid squaring here; convert the left side to one sine and use sin A = sin B.
2. Take sqrt x as a non-negative quantity, so only n >= 0 is allowed.