JEE Mathematics Trigonometric Equations MCQs Set 04

Answer: (a) \( x=(2n+1)\frac{\pi}{2}\ \text{or}\ x=n\pi\pm\frac{\pi}{6} \)

Explanation:
1. Do not cancel \( \cos x \); factorise: \( \cos x\,(2\cos 2x-1)=0 \).
2. \( \cos x=0\Rightarrow x=(2n+1)\frac{\pi}{2} \).
3. \( \cos 2x=\frac12=\cos\frac{\pi}{3}\Rightarrow 2x=2n\pi\pm\frac{\pi}{3}\Rightarrow x=n\pi\pm\frac{\pi}{6} \).

Teacher's Note:
1. Never divide by a trigonometric factor, you lose its roots. Move everything to one side and factorise.
2. For cos 2x = cos α the general solution is 2x = 2nπ ± α.

Answer: (a) \( x=(2n+1)\pi\ \text{or}\ x=n\pi+(-1)^n\frac{\pi}{6} \)

Explanation:
1. Write \( \sin^2x=(1-\cos x)(1+\cos x) \) and take \( (1+\cos x) \) common.
2. \( (1+\cos x)\,[2\sin x-\cos x-(1-\cos x)]=0\Rightarrow(1+\cos x)(2\sin x-1)=0 \).
3. \( \cos x=-1\Rightarrow x=(2n+1)\pi \); \( \sin x=\frac12\Rightarrow x=n\pi+(-1)^n\frac{\pi}{6} \).

Teacher's Note:
1. Spot the common factor (1 + cos x) after replacing sin²x by (1 − cos x)(1 + cos x).
2. sin x = sin α gives x = nπ + (−1)ⁿ α.

Answer: (a) \( x=2n\pi\pm\cos^{-1}\frac13 \)

Explanation:
1. Factorise: \( (3\cos x-1)(\cos x-3)=0 \).
2. \( \cos x=3 \) is impossible since \( |\cos x|\le1 \).
3. \( \cos x=\frac13\Rightarrow x=2n\pi\pm\cos^{-1}\frac13 \).

Teacher's Note:
1. Treat it as a quadratic in cos x and reject any value outside [−1, 1].
2. cos x = cos α gives x = 2nπ ± α.

Answer: (a) \( x=n\pi\pm\frac{\pi}{4} \)

Explanation:
1. Put \( s=\sin^2x \): \( 2\cdot4s(1-s)+6s=5\Rightarrow 8s^2-14s+5=0 \).
2. \( s=\frac{14\pm6}{16}=\frac54\ \text{or}\ \frac12 \); \( \frac54 \) is rejected.
3. \( \sin^2x=\frac12=\sin^2\frac{\pi}{4}\Rightarrow x=n\pi\pm\frac{\pi}{4} \).

Teacher's Note:
1. Express everything in sin²x using sin 2x = 2 sin x cos x.
2. sin²x = sin²α gives x = nπ ± α.

Answer: (a) \( 5050\pi \)

Explanation:
1. Put \( c=\cos2x \): \( 2c^2-1+6=7c\Rightarrow2c^2-7c+5=0\Rightarrow c=1\ \text{or}\ \frac52 \).
2. \( \cos2x=1\Rightarrow x=n\pi \), so the solutions in the interval are \( 0,\pi,2\pi,\dots,100\pi \).
3. Sum \( =\pi(0+1+\dots+100)=5050\pi \).

Teacher's Note:
1. Reject c = 5/2 because cos 2x cannot exceed 1.
2. Use the formula for the sum of the first n natural numbers to add the multiples of π.

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

Explanation:
1. Use \( \sec^2x=1+\tan^2x \) with \( t=\tan^2x \): \( 5t^2-(1+t)^2=29\Rightarrow2t^2-t-15=0 \).
2. \( t=3\ \text{or}\ -\frac52 \); negative value rejected.
3. \( \tan^2x=3=\tan^2\frac{\pi}{3}\Rightarrow x=n\pi\pm\frac{\pi}{3} \).

Teacher's Note:
1. Convert sec² to tan² and solve the resulting quadratic in tan² x.
2. tan²x is never negative.

Answer: (a) \( x=2n\pi+\frac{\pi}{4} \)

Explanation:
1. \( \sin x+\cos x=\sqrt2\sin\left(x+\frac{\pi}{4}\right) \).
2. \( \sin\left(x+\frac{\pi}{4}\right)=1\Rightarrow x+\frac{\pi}{4}=2n\pi+\frac{\pi}{2} \).

Teacher's Note:
1. Write a sin x + b cos x as R sin(x + φ) with R = √(a² + b²).
2. Here R = √2, so the maximum value √2 is attained.

Answer: (a) \( x=2n\pi\ \text{or}\ x=2n\pi+\frac{\pi}{4} \)

Explanation:
1. Multiply by \( \cos x \): \( 1-\cos x=(\sqrt2-1)\sin x \), i.e. \( 2\sin^2\frac x2=(\sqrt2-1)\,2\sin\frac x2\cos\frac x2 \).
2. \( \sin\frac x2=0\Rightarrow x=2n\pi \).
3. \( \tan\frac x2=\sqrt2-1=\tan\frac{\pi}{8}\Rightarrow x=2n\pi+\frac{\pi}{4} \).

Teacher's Note:
1. Half-angle forms 1 − cos x = 2 sin²(x/2) and sin x = 2 sin(x/2) cos(x/2) simplify such equations.
2. tan(π/8) = √2 − 1 is worth remembering.

Answer: (a) \( x=\frac{m\pi}{2}+\frac{\pi}{16}\ \text{or}\ x=\frac{(8m+3)\pi}{24} \)

Explanation:
1. \( \sin\left(x+\frac{\pi}{4}\right)=\sin5x \).
2. If \( 5x=2m\pi+x+\frac{\pi}{4} \): \( x=\frac{m\pi}{2}+\frac{\pi}{16} \).
3. If \( 5x=(2m+1)\pi-x-\frac{\pi}{4} \): \( 6x=(2m+1)\pi-\frac{\pi}{4}\Rightarrow x=\frac{(8m+3)\pi}{24} \).

Teacher's Note:
1. Reduce the left side to √2 sin(x + π/4) and use sin A = sin B ⇒ A = 2mπ + B or A = (2m+1)π − B.
2. Check one root numerically, for example x = π/16.

Answer: (a) \( x=\frac{(12m-1)\pi}{84}\ \text{or}\ x=\frac{(12m+7)\pi}{48} \)

Explanation:
1. \( \cos3x+\sqrt3\sin3x=2\sin\left(3x+\frac{\pi}{6}\right) \), so \( \sin11x=-\sin\left(3x+\frac{\pi}{6}\right)=\sin\left(-3x-\frac{\pi}{6}\right) \).
2. First family: \( 11x=2m\pi-3x-\frac{\pi}{6}\Rightarrow x=\frac{(12m-1)\pi}{84} \).
3. Second family: \( 11x=(2m+1)\pi+3x+\frac{\pi}{6}\Rightarrow x=\frac{(12m+7)\pi}{48} \).

Teacher's Note:
1. Combine cos 3x and √3 sin 3x into 2 sin(3x + π/6) first.
2. Then use sin A = sin B for the two families.

Answer: (a) \( \sin\alpha+\sin\beta=\frac{2bc}{a^2+b^2} \) and \( \sin\alpha\sin\beta=\frac{c^2-a^2}{a^2+b^2} \)

Explanation:
1. Write \( a\cos\theta=c-b\sin\theta \) and square: \( (a^2+b^2)\sin^2\theta-2bc\sin\theta+c^2-a^2=0 \).
2. The values \( \sin\alpha,\sin\beta \) are the two roots of this quadratic.
3. Sum of roots \( =\frac{2bc}{a^2+b^2} \), product \( =\frac{c^2-a^2}{a^2+b^2} \). (Similarly \( \tan\frac\alpha2+\tan\frac\beta2=\frac{2b}{a+c} \) and \( \tan\frac\alpha2\tan\frac\beta2=\frac{c-a}{c+a} \).)

Teacher's Note:
1. Eliminate cos θ to get a quadratic in sin θ, then use sum and product of roots.
2. Using the half-angle substitution t = tan(θ/2) gives the tangent results instead.

Answer: (a) \( x=(2n+1)\pi\ \text{or}\ x=2n\pi-\frac{\pi}{2} \)

Explanation:
1. Let \( s=\sin x+\cos x \), \( p=\sin x\cos x=\frac{s^2-1}{2} \). Then \( \sin^3x+\cos^3x=s(1-p) \).
2. \( 1+s(1-p)=3p\Rightarrow s^3+3s^2-3s-5=0\Rightarrow(s+1)(s^2+2s-5)=0 \).
3. \( s^2+2s-5=0 \) gives \( s=-1\pm\sqrt6 \), outside \( [-\sqrt2,\sqrt2] \). So \( s=-1 \).
4. \( \sin x+\cos x=-1\Rightarrow x=(2n+1)\pi\ \text{or}\ x=2n\pi-\frac{\pi}{2} \).

Teacher's Note:
1. Use the symmetric substitution s = sin x + cos x, p = sin x cos x = (s² − 1)/2.
2. Check candidate roots against |s| ≤ √2.

Answer: (a) \( x=2n\pi\ \text{or}\ x=2n\pi+\frac{2\pi}{3} \)

Explanation:
1. Let \( a=\sqrt3\sin x,\ b=\cos x \): the equation is \( a^3+b^3+(-1)^3-3ab(-1)=0 \).
2. \( (a+b-1)(a^2+b^2+1-ab+a+b)=0 \). The second factor vanishes only if \( a=b=-1 \), which is impossible here.
3. \( \sqrt3\sin x+\cos x=1\Rightarrow\sin\left(x+\frac{\pi}{6}\right)=\frac12 \), so \( x=2n\pi \) or \( x=2n\pi+\frac{2\pi}{3} \).

Teacher's Note:
1. Recognise the identity a³ + b³ + c³ − 3abc = (a + b + c)(…) with c = −1.
2. The second factor is a sum of squares and is zero only when a = b = c.

Answer: (a) \( x=\frac{n\pi}{3} \)

Explanation:
1. Sum to product: \( 2\sin3x\cos2x=2\sin3x\cos x \).
2. \( \sin3x\,(\cos2x-\cos x)=0 \).
3. \( \sin3x=0\Rightarrow x=\frac{n\pi}{3} \); \( \cos2x=\cos x\Rightarrow x=\frac{2n\pi}{3} \) (already included).

Teacher's Note:
1. Group the terms with equal sums of angles: (x, 5x) and (2x, 4x) both have mean 3x and 3x respectively.
2. The second family is a subset of the first.

Answer: (a) No solution

Explanation:
1. \( \sin x-\sin3x=-2\cos2x\sin x \), so the equation is \( 2\sin x(2\cos x-\cos2x)=3 \).
2. For \( x\in[0,\pi] \) the left-hand side \( \sin x+2\sin2x-\sin3x \) has maximum about 2.73, which is less than 3.
3. Hence there is no solution.

Teacher's Note:
1. When a sum of bounded terms cannot reach the right side, there are no solutions.
2. A quick graph or the maximum of the left side settles it.

Answer: (b) 28

Explanation:
1. \( \tan A=\cot B\Rightarrow\cos(A+B)=0\Rightarrow5\pi(\cos\alpha+\sin\alpha)=(2k+1)\frac{\pi}{2} \).
2. \( \cos\alpha+\sin\alpha=\frac{2k+1}{10} \) and \( |\cos\alpha+\sin\alpha|\le\sqrt2 \), so \( 2k+1=\pm1,\pm3,\dots,\pm13 \): 14 values.
3. Each gives two values of \( \alpha \) in \( [0,2\pi] \), and none makes tan or cot undefined. Total \( 14\times2=28 \).

Teacher's Note:
1. Convert tan A = cot B to cos(A + B) = 0.
2. Count integer values of k that keep the right side within [−√2, √2].

Answer: (a) \( x=(2n+1)\frac{\pi}{14} \)

Explanation:
1. Since \( 8\cos x\cos2x\cos4x=\frac{\sin8x}{\sin x} \) (repeated double angle), the equation is \( \sin6x=\sin8x \) with \( \sin x\ne0 \).
2. \( 8x=2m\pi+6x\Rightarrow x=m\pi \) is rejected since \( \sin x=0 \).
3. \( 8x=(2m+1)\pi-6x\Rightarrow x=(2m+1)\frac{\pi}{14} \) (never a multiple of π, so valid).

Teacher's Note:
1. Use sin 2θ = 2 sin θ cos θ repeatedly: sin 8x = 8 sin x cos x cos 2x cos 4x.
2. Always exclude the values that make a denominator zero.

Answer: (a) \( x=\frac{n\pi}{2}+\frac{\pi}{8} \)

Explanation:
1. Let \( p=\sin2x\cos2x=\frac12\sin4x \). Then \( \sin^42x+\cos^42x=1-2p^2 \).
2. \( 1-2p^2=p\Rightarrow2p^2+p-1=0\Rightarrow p=\frac12\ \text{or}\ -1 \).
3. \( |p|\le\frac12 \), so \( p=\frac12\Rightarrow\sin4x=1\Rightarrow4x=2n\pi+\frac{\pi}{2}\Rightarrow x=\frac{n\pi}{2}+\frac{\pi}{8} \).

Teacher's Note:
1. Use a⁴ + b⁴ = (a² + b²)² − 2a²b².
2. Reject any root outside the range of sin 4x / 2.

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

Explanation:
1. \( \sqrt{1-\cos x}=\sqrt2\left|\sin\frac x2\right| \), and the right side needs \( \sin x\ge0 \).
2. Squaring: \( 1-\cos x=\sin^2x\Rightarrow\cos x(\cos x-1)=0\Rightarrow\cos x=0\ \text{or}\ 1 \).
3. \( \cos x=1\Rightarrow x=2n\pi \) (valid). \( \cos x=0 \): \( \sin x=1 \) gives \( x=2n\pi+\frac{\pi}{2} \) (valid), while \( \sin x=-1 \) is rejected.

Teacher's Note:
1. After squaring always test the roots in the original equation.
2. The right side must be non-negative since the left is a square root.

Answer: (a) \( x=(m+n)\pi\pm\frac{\pi}{6},\ y=(m-n)\pi\mp\frac{\pi}{6} \)

Explanation:
1. \( \cos(x+y)=1\Rightarrow x+y=2m\pi \); \( \cos(x-y)=\frac12\Rightarrow x-y=2n\pi\pm\frac{\pi}{3} \).
2. Adding and halving: \( x=(m+n)\pi\pm\frac{\pi}{6} \).
3. Subtracting and halving: \( y=(m-n)\pi\mp\frac{\pi}{6} \).

Teacher's Note:
1. Solve for x + y and x − y separately, then add and subtract.
2. Keep the ± signs consistent between x and y.

Answer: (a) \( \frac{1}{243} \)

Explanation:
1. \( 27^{\cos2x}81^{\sin2x}=3^{3\cos2x+4\sin2x} \).
2. \( 3\cos2x+4\sin2x \) ranges over \( [-5,5] \).
3. Minimum \( =3^{-5}=\frac{1}{243} \).

Teacher's Note:
1. Write both bases as powers of 3, then minimise the exponent.
2. a cos θ + b sin θ has minimum −√(a² + b²).

Answer: (c) 7

Explanation:
1. \( (\cos x+\cos4x)+(\cos2x+\cos3x)=2\cos\frac{5x}{2}\left(\cos\frac{3x}{2}+\cos\frac x2\right)=4\cos\frac{5x}{2}\cos x\cos\frac x2 \).
2. \( \cos\frac{5x}{2}=0\Rightarrow x=\frac{\pi}{5},\frac{3\pi}{5},\pi,\frac{7\pi}{5},\frac{9\pi}{5} \); \( \cos x=0\Rightarrow x=\frac{\pi}{2},\frac{3\pi}{2} \); \( \cos\frac x2=0\Rightarrow x=\pi \) (already counted).
3. Total distinct values: 7.

Teacher's Note:
1. Pair the cosines whose angle sums are equal and convert to products.
2. Do not count the repeated root x = π twice.

Answer: (a) \( n\pi+\frac{\pi}{8}\lt x\lt n\pi+\frac{\pi}{4} \)

Explanation:
1. Divide by \( \cos^2x\gt 0 \) (cos x = 0 gives 0 > √2, false), with \( t=\tan x \): \( 2t\gt \sqrt2t^2+2-\sqrt2 \).
2. \( \sqrt2t^2-2t+(2-\sqrt2)\lt 0\Rightarrow(t-1)(\sqrt2t-(2-\sqrt2))\lt 0 \), with roots \( t=1 \) and \( t=\sqrt2-1=\tan\frac{\pi}{8} \).
3. \( \sqrt2-1\lt \tan x\lt 1\Rightarrow n\pi+\frac{\pi}{8}\lt x\lt n\pi+\frac{\pi}{4} \).

Teacher's Note:
1. Divide by cos²x to turn a homogeneous inequality into a quadratic inequality in tan x.
2. Check the case cos x = 0 separately.

Answer: (a) \( n\pi+\frac{\pi}{4}\lt x\lt n\pi+\frac{\pi}{3} \)

Explanation:
1. Factorise: \( (\tan x-1)(\tan x-\sqrt3)\lt 0 \).
2. \( 1\lt \tan x\lt \sqrt3 \).
3. \( n\pi+\frac{\pi}{4}\lt x\lt n\pi+\frac{\pi}{3} \).

Teacher's Note:
1. Because tan has period π, the answer repeats with nπ, not 2nπ.
2. On each branch tan is increasing, so the inequality on tan x maps directly to x.