JEE Mathematics Trigonometric Equations MCQs Set 06

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

Explanation:
1. \( 1-\sin2x=(\cos x-\sin x)^2 \), so \( (\cos x-\sin x)(\cos x-\sin x-1)=0 \).
2. \( \cos x=\sin x\Rightarrow\tan x=1\Rightarrow x=n\pi+\frac{\pi}{4} \).
3. \( \cos x-\sin x=1\Rightarrow\cos\left(x+\frac{\pi}{4}\right)=\frac{1}{\sqrt2}\Rightarrow x=2n\pi \) or \( x=2n\pi-\frac{\pi}{2} \).

Teacher's Note:
1. The left side is a perfect square of cos x − sin x, so the equation factorises at once.
2. Use the R-formula for the second factor.

Answer: (c) \( x = n\pi \) or \( x = n\pi+(-1)^n\frac{\pi}{10} \) or \( x = n\pi-(-1)^n\frac{3\pi}{10} \)

Explanation:
1. \( \sin x\left(4\cos^2x-2\sin x-3\right)=0 \).
2. \( \sin x=0\Rightarrow x=n\pi \). Otherwise \( 4-4\sin^2x-2\sin x-3=0\Rightarrow4\sin^2x+2\sin x-1=0 \).
3. \( \sin x=\frac{-1\pm\sqrt5}{4}=\sin\frac{\pi}{10} \) or \( \sin\left(-\frac{3\pi}{10}\right) \): \( x=n\pi+(-1)^n\frac{\pi}{10} \) or \( x=n\pi-(-1)^n\frac{3\pi}{10} \).

Teacher's Note:
1. Factor out sin x instead of cancelling it, otherwise the family x = nπ is lost.

Answer: (b) \( x = n\pi \)

Explanation:
1. \( 2\cos^2\theta-1=\cos^2\theta\Rightarrow\cos^2\theta=1 \).
2. \( \sin\theta=0 \).
3. \( \theta=n\pi \).

Teacher's Note:
1. cos²θ = 1 means cos θ = ±1, which is the same as sin θ = 0.
2. So the solution is nπ rather than 2nπ.

Answer: (c) \( \frac{2\pi}{p+q} \) and \( \frac{2\pi}{p-q} \)

Explanation:
1. \( 2\cos\frac{(p+q)\theta}{2}\cos\frac{(p-q)\theta}{2}=0 \).
2. \( \frac{(p+q)\theta}{2}=(2k+1)\frac{\pi}{2}\Rightarrow\theta=\frac{(2k+1)\pi}{p+q} \): an AP with common difference \( \frac{2\pi}{p+q} \).
3. \( \frac{(p-q)\theta}{2}=(2k+1)\frac{\pi}{2}\Rightarrow\theta=\frac{(2k+1)\pi}{p-q} \): common difference \( \frac{2\pi}{p-q} \).

Teacher's Note:
1. Increasing k by 1 changes (2k + 1) by 2, which gives the common difference 2π divided by the denominator.

Answer: (d) \( x=\frac{(8m+1)^2\pi^2}{4} \) or \( x=\frac{(8m+3)^2\pi^2}{36} \)

Explanation:
1. \( \sin\frac{\sqrt x}{2}+\cos\frac{\sqrt x}{2}=\sqrt2\sin\left(\frac{\sqrt x}{2}+\frac{\pi}{4}\right) \), so \( \sin\left(\frac{\sqrt x}{2}+\frac{\pi}{4}\right)=\sin\sqrt x \). Let \( s=\sqrt x\ge0 \).
2. Case 1: \( s=2m\pi+\frac{s}{2}+\frac{\pi}{4}\Rightarrow s=\frac{(8m+1)\pi}{2} \), so \( x=\frac{(8m+1)^2\pi^2}{4} \).
3. Case 2: \( s=(2m+1)\pi-\frac{s}{2}-\frac{\pi}{4}\Rightarrow\frac{3s}{2}=\frac{(8m+3)\pi}{4}\Rightarrow s=\frac{(8m+3)\pi}{6} \), so \( x=\frac{(8m+3)^2\pi^2}{36} \). Since \( s\ge0 \), \( m\ge0 \) in both cases.

Teacher's Note:
1. Put s = √x and solve sin A = sin B in s, remembering s ≥ 0.
2. Squaring s at the end gives the values of x.

Answer: (b) \( x = \frac{n\pi}{3} \) or \( x = n\pi\pm\tan^{-1}\frac{1}{\sqrt2} \)

Explanation:
1. \( \tan\theta+\tan2\theta=\frac{\sin3\theta}{\cos\theta\cos2\theta} \), so \( \sin3\theta\left[\frac{1}{\cos\theta\cos2\theta}+\frac{1}{\cos3\theta}\right]=0 \), i.e. \( \sin3\theta(\cos3\theta+\cos\theta\cos2\theta)=0 \).
2. \( \sin3\theta=0\Rightarrow\theta=\frac{n\pi}{3} \). Also \( \cos3\theta+\cos\theta\cos2\theta=6c^3-4c=2c(3c^2-2)=0 \) with \( c=\cos\theta \).
3. \( c=0 \) is rejected (tan θ undefined). \( \cos^2\theta=\frac23\Rightarrow\tan^2\theta=\frac12\Rightarrow\theta=n\pi\pm\tan^{-1}\frac{1}{\sqrt2} \).

Teacher's Note:
1. Combine the first two tangents into sin 3θ over cos θ cos 2θ so that sin 3θ factors out.
2. Use cos 3θ = 4cos³θ − 3cos θ for the remaining factor.

Answer: (b) \( x = 2n\pi+\frac{2\pi}{3} \)

Explanation:
1. Multiply by \( \sin\theta \): \( 1=\cos\theta+\sqrt3\sin\theta\Rightarrow\sin\left(\theta+\frac{\pi}{6}\right)=\frac12 \).
2. \( \theta+\frac{\pi}{6}=n\pi+(-1)^n\frac{\pi}{6} \): even \( n \) gives \( \theta=2m\pi \); odd \( n \) gives \( \theta=2m\pi+\frac{2\pi}{3} \).
3. But \( \theta=2m\pi \) makes \( \sin\theta=0 \) and \( \csc\theta \) undefined, so only \( \theta=2n\pi+\frac{2\pi}{3} \) remains. Check: \( \csc\frac{2\pi}{3}=\frac{2}{\sqrt3}=\cot\frac{2\pi}{3}+\sqrt3=-\frac{1}{\sqrt3}+\sqrt3 \) ✓.

Teacher's Note:
1. Multiplying by sin θ can introduce the extraneous roots where sin θ = 0.
2. Always test them in the original equation.

Answer: (b) 2, 1, 1, 0

Explanation:
1. \( \sin x=\frac57 \): positive, so \( x \) and \( 180^\circ-x \) both lie in \( (0^\circ,180^\circ) \): 2 values.
2. \( \cos x \) is one-to-one on \( (0^\circ,180^\circ) \), so (ii) and (iii) each have exactly 1 value.
3. \( \cot x=-7\Rightarrow\tan x=-\frac17 \), which contradicts \( \tan x=\frac23 \). No angle satisfies both: 0 values.

Teacher's Note:
1. Sine takes each value in (0, 1) twice in (0°, 180°), while cosine and tangent take each value once.
2. Two conditions on the same angle may be inconsistent.

Answer: (b) \( \frac{1}{2\sqrt2} \)

Explanation:
1. From \( \sin A=\cos B \): \( \pi\cos\theta=\frac{\pi}{2}\mp\pi\sin\theta+2k\pi \) (both signs), so \( \cos\theta\pm\sin\theta=\frac12+2k \).
2. Since \( |\cos\theta\pm\sin\theta|\le\sqrt2\lt\frac32 \), only \( k=0 \) is possible: \( \cos\theta\pm\sin\theta=\frac12 \).
3. \( \cos\theta\pm\sin\theta=\sqrt2\cos\left(\theta\mp\frac{\pi}{4}\right) \), so \( \cos\left(\theta\pm\frac{\pi}{4}\right)=\frac{1}{2\sqrt2} \).

Teacher's Note:
1. Convert sin A = cos B into a linear relation between A and B and use the bound |cos θ ± sin θ| ≤ √2 to fix the integer.
2. The R-formula finishes the problem.

Answer: (b) \( x = \frac{(2n+1)\pi}{6} \) or \( x = \frac{2n\pi}{3}\pm\frac{2\pi}{9} \)

Explanation:
1. Let \( c=\cos3x \): \( \cos9x=4c^3-3c \), \( \cos6x=2c^2-1 \). Then \( 4c^3-3c-4c^2+2=2 \).
2. \( 4c^3-4c^2-3c=c(2c+1)(2c-3)=0 \), so \( c=0 \), \( c=-\frac12 \) or \( c=\frac32 \) (rejected).
3. \( \cos3x=0\Rightarrow x=\frac{(2n+1)\pi}{6} \); \( \cos3x=-\frac12\Rightarrow3x=2n\pi\pm\frac{2\pi}{3}\Rightarrow x=\frac{2n\pi}{3}\pm\frac{2\pi}{9} \).

Teacher's Note:
1. Express cos 9x and cos 6x in terms of cos 3x, then solve the cubic.
2. The root 3/2 is rejected because cosine cannot exceed 1.

Answer: (d) 5025

Explanation:
1. \( \tan\pi x=-1\Rightarrow\pi x=n\pi-\frac{\pi}{4}\Rightarrow x=n-\frac14 \).
2. In \( [0,100] \): \( n=1,2,\ldots,100 \) gives \( x=\frac34,\frac74,\ldots,\frac{399}{4} \) (100 solutions); \( n=101 \) gives \( 100.75\gt100 \).
3. \( \sum_{n=1}^{100}\left(n-\frac14\right)=5050-25=5025 \).

Teacher's Note:
1. The solutions form an arithmetic progression with common difference 1.
2. Count the terms carefully at both ends of the interval.

Answer: (d) \( x+y=2n\pi+\frac{\pi}{2} \), \( x-y=m\pi+(-1)^{m+1}\sin^{-1}\frac35 \)

Explanation:
1. \( 4\tan x=\tan y\Rightarrow4\sin x\cos y=\cos x\sin y \) and \( 5\sin x\cos y=1\Rightarrow\sin x\cos y=\frac15 \), so \( \cos x\sin y=\frac45 \).
2. \( \sin(x+y)=\frac15+\frac45=1\Rightarrow x+y=2n\pi+\frac{\pi}{2} \). \( \sin(x-y)=\frac15-\frac45=-\frac35 \).
3. \( x-y=m\pi+(-1)^m\left(-\sin^{-1}\frac35\right)=m\pi+(-1)^{m+1}\sin^{-1}\frac35 \).

Teacher's Note:
1. Use the two products sin x cos y and cos x sin y to build sin(x + y) and sin(x − y).
2. Solve for the sum and the difference of the angles.

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

Explanation:
1. The logarithms need \( \cos x+\sin x\gt0 \) and \( \cos x-\sin x\gt0 \). The equation becomes \( \sqrt3(\cos x+\sin x)-(\cos x-\sin x)=\sqrt2 \), i.e. \( (\sqrt3-1)\cos x+(\sqrt3+1)\sin x=\sqrt2 \).
2. Divide by \( 2\sqrt2 \): \( \cos75^\circ\cos x+\sin75^\circ\sin x=\frac12\Rightarrow\cos\left(x-\frac{5\pi}{12}\right)=\frac12 \), so \( x=2n\pi+\frac{3\pi}{4} \) or \( 2n\pi+\frac{\pi}{12} \).
3. For \( x=\frac{3\pi}{4} \), \( \cos x-\sin x=-\sqrt2\lt0 \), rejected. For \( x=\frac{\pi}{12} \) both conditions hold. So \( x=2n\pi+\frac{\pi}{12} \).

Teacher's Note:
1. Remove the logarithms using a^(log_a y) = y, but keep the conditions y > 0 for the domain.
2. Here one of the two roots violates cos x − sin x > 0.

Answer: (d) \( x = n\pi\pm\frac{\pi}{6} \)

Explanation:
1. \( \sec^2\theta=1+\tan^2\theta \): \( 3(1+2\tan^2\theta)=5\Rightarrow\tan^2\theta=\frac13 \).
2. \( \tan^2\theta=\tan^2\frac{\pi}{6} \).
3. \( \theta=n\pi\pm\frac{\pi}{6} \).

Teacher's Note:
1. Reduce to a single function and use the tan²θ = tan²α rule, which gives nπ ± α.

Answer: (d) \( x = \frac{n\pi}{3} \) or \( x = \frac{n\pi}{2}\pm\frac{\pi}{12} \)

Explanation:
1. \( \sin7\theta-\sin\theta=2\cos4\theta\sin3\theta \), so \( \sin3\theta(2\cos4\theta-1)=0 \).
2. \( \sin3\theta=0\Rightarrow\theta=\frac{n\pi}{3} \).
3. \( \cos4\theta=\frac12\Rightarrow4\theta=2n\pi\pm\frac{\pi}{3}\Rightarrow\theta=\frac{n\pi}{2}\pm\frac{\pi}{12} \).

Teacher's Note:
1. Move sin θ to the left so that the sin 7θ − sin θ pair gives a factor sin 3θ, which is already on the right.

Answer: (c) \( n+\frac14 \)

Explanation:
1. \( \sin(\pi\cot\theta)=\cos(\pi\tan\theta)\Rightarrow\pi\cot\theta=\frac{\pi}{2}\mp\pi\tan\theta+2n\pi \).
2. \( \cot\theta+\tan\theta=2n+\frac12 \) or \( \cot\theta-\tan\theta=2n+\frac12 \).
3. \( \cot\theta+\tan\theta=\frac{2}{\sin2\theta} \) and \( \cot\theta-\tan\theta=2\cot2\theta \), so \( \csc2\theta=n+\frac14 \) or \( \cot2\theta=n+\frac14 \).

Teacher's Note:
1. Use cot θ + tan θ = 2 cosec 2θ and cot θ − tan θ = 2 cot 2θ after splitting the two cases.
2. Dividing by 2 turns 2n + ½ into n + ¼.

Answer: (c) \( x = n\pi+(-1)^n\frac{\pi}{6} \)

Explanation:
1. \( 1-\sin^2\theta-\sin\theta-\frac14=0\Rightarrow4\sin^2\theta+4\sin\theta-3=0 \).
2. \( (2\sin\theta-1)(2\sin\theta+3)=0 \). The root \( \sin\theta=-\frac32 \) is impossible.
3. \( \sin\theta=\frac12\Rightarrow\theta=n\pi+(-1)^n\frac{\pi}{6} \).

Teacher's Note:
1. Replace cos²θ by 1 − sin²θ to obtain a quadratic in sin θ.
2. Reject any root outside [−1, 1].

Answer: (a) \( x = 2n\pi\pm\frac{\pi}{4}+\frac{\pi}{6} \)

Explanation:
1. \( \sqrt3\cos\theta+\sin\theta=2\cos\left(\theta-\frac{\pi}{6}\right) \).
2. \( \cos\left(\theta-\frac{\pi}{6}\right)=\frac{1}{\sqrt2}\Rightarrow\theta-\frac{\pi}{6}=2n\pi\pm\frac{\pi}{4} \).
3. \( \theta=2n\pi\pm\frac{\pi}{4}+\frac{\pi}{6} \).

Teacher's Note:
1. Here R = √(3 + 1) = 2 and the phase angle is π/6, because cos α = √3/2 and sin α = 1/2.

Answer: (b) \( \frac{\pi}{7},\frac{5\pi}{7},\pi,\frac{9\pi}{7},\frac{13\pi}{7} \)

Explanation:
1. \( \cos3x+\cos2x=2\cos\frac{5x}{2}\cos\frac{x}{2} \) and \( \sin\frac{3x}{2}+\sin\frac{x}{2}=2\sin x\cos\frac{x}{2} \), so \( 2\cos\frac{x}{2}\left(\cos\frac{5x}{2}-\sin x\right)=0 \).
2. \( \cos\frac{x}{2}=0\Rightarrow x=\pi \). \( \cos\frac{5x}{2}=\cos\left(\frac{\pi}{2}-x\right)\Rightarrow\frac{5x}{2}=2n\pi\pm\left(\frac{\pi}{2}-x\right) \).
3. Plus: \( x=\frac{(4n+1)\pi}{7} \) gives \( \frac{\pi}{7},\frac{5\pi}{7},\frac{9\pi}{7},\frac{13\pi}{7} \). Minus: \( x=\frac{(4n-1)\pi}{3} \) gives only \( \pi \) in the interval. So the five values are listed in the first option.

Teacher's Note:
1. Use sum-to-product formulas on both sides; cos(x/2) is a common factor.
2. Count the values of each family inside (0, 2π).

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

Explanation:
1. Use \( \cos2x=1-2\sin^2x \): \( 2\sin^2x-3\sin x+1=0 \).
2. \( (2\sin x-1)(\sin x-1)=0\Rightarrow\sin x=\frac12 \) or \( \sin x=1 \).
3. \( x=n\pi+(-1)^n\frac{\pi}{6} \) or \( x=2n\pi+\frac{\pi}{2} \).

Teacher's Note:
1. Convert cos 2x into sin x so that the equation becomes a quadratic in one function.
2. Both roots of the quadratic lie in [−1, 1], so both are valid.

Answer: (c) \( -2,-1,0,1 \)

Explanation:
1. Let \( t=\tan x \). Squaring: \( 13-18t=36t^2-36t+9\Rightarrow18t^2-9t-2=0 \), so \( t=\frac23 \) or \( t=-\frac16 \).
2. The right side must be non-negative: \( 6t-3\ge0\Rightarrow t\ge\frac12 \). So \( t=\frac23 \) (check: \( \sqrt{13-12}=1=6\cdot\frac23-3 \)).
3. \( x=\tan^{-1}\frac23+k\pi\approx0.588+k\pi \) lies in \( (-2\pi,2\pi) \) for \( k=-2,-1,0,1 \) (for \( k=2 \), \( x\approx6.87\gt2\pi \)).

Teacher's Note:
1. A square-root equation needs the condition that the right-hand side is non-negative, which removes t = −1/6.
2. Then count the solutions in the interval.

Answer: (c) \( x = n\pi+\frac{\pi}{4} \) or \( x = n\pi+\tan^{-1}\frac12 \)

Explanation:
1. Write \( 1=\sin^2\theta+\cos^2\theta \): \( 2\sin^2\theta+\cos^2\theta=3\sin\theta\cos\theta \). Divide by \( \cos^2\theta \) (\( \cos\theta=0 \) is not a root): \( 2\tan^2\theta-3\tan\theta+1=0 \).
2. \( (2\tan\theta-1)(\tan\theta-1)=0\Rightarrow\tan\theta=\frac12 \) or \( 1 \).
3. \( \theta=n\pi+\tan^{-1}\frac12 \) or \( \theta=n\pi+\frac{\pi}{4} \).

Teacher's Note:
1. A constant in a homogeneous quadratic form is written as sin²θ + cos²θ.
2. Then divide by cos²θ to get a quadratic in tan θ.

Answer: (d) \( x = \frac{2n\pi}{5} \) or \( x = \frac{(2n+1)\pi}{2} \) or \( x = (2n+1)\pi \)

Explanation:
1. \( (\sin4x+\sin x)+(\sin3x+\sin2x)=2\sin\frac{5x}{2}\left[\cos\frac{3x}{2}+\cos\frac{x}{2}\right]=4\sin\frac{5x}{2}\cos x\cos\frac{x}{2} \).
2. \( \sin\frac{5x}{2}=0\Rightarrow x=\frac{2n\pi}{5} \); \( \cos x=0\Rightarrow x=\frac{(2n+1)\pi}{2} \).
3. \( \cos\frac{x}{2}=0\Rightarrow x=(2n+1)\pi \).

Teacher's Note:
1. Pair the terms so that the half-sum 5x/2 appears in both pairs.
2. The final factorisation has three factors and so three families of roots.

Answer: (a) \( x = \frac{n\pi}{3}+\frac{\pi}{9} \)

Explanation:
1. \( \tan\theta+\tan2\theta=\sqrt3(1-\tan\theta\tan2\theta)\Rightarrow\frac{\tan\theta+\tan2\theta}{1-\tan\theta\tan2\theta}=\sqrt3 \).
2. That is \( \tan3\theta=\sqrt3 \).
3. \( 3\theta=n\pi+\frac{\pi}{3}\Rightarrow\theta=\frac{n\pi}{3}+\frac{\pi}{9} \).

Teacher's Note:
1. Recognise the tan(A + B) formula in disguise.
2. The general solution of tan 3θ = √3 is then divided by 3.

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

Explanation:
1. Use half angles: \( 1-\sin\theta=\left(\cos\frac{\theta}{2}-\sin\frac{\theta}{2}\right)^2 \) and \( \cos\theta=\left(\cos\frac{\theta}{2}-\sin\frac{\theta}{2}\right)\left(\cos\frac{\theta}{2}+\sin\frac{\theta}{2}\right) \).
2. Factor out \( \left(c-s\right) \) with \( c=\cos\frac{\theta}{2} \), \( s=\sin\frac{\theta}{2} \): either \( c=s \), i.e. \( \tan\frac{\theta}{2}=1\Rightarrow\theta=2n\pi+\frac{\pi}{2} \),
3. or \( (2+\sqrt3)(c+s)=c-s\Rightarrow\tan\frac{\theta}{2}=-\frac{1+\sqrt3}{3+\sqrt3}=-\frac{1}{\sqrt3}\Rightarrow\theta=2n\pi-\frac{\pi}{3} \).

Teacher's Note:
1. Half-angle forms turn 1 − sin θ into a perfect square, so a common factor appears.
2. Check θ = −π/3: cos = ½ and 1 − sin = 1 + √3/2, and (2 + √3)/2 equals that.

Answer: (a) \( 3\pi \)

Explanation:
1. \( \sin5\theta+\sin\theta=2\sin3\theta\cos2\theta \), so \( \sin3\theta(2\cos2\theta-1)=0 \).
2. \( \sin3\theta=0 \): \( \theta=0,\frac{\pi}{3},\frac{2\pi}{3},\pi \). \( \cos2\theta=\frac12 \): \( 2\theta=\frac{\pi}{3},\frac{5\pi}{3} \), so \( \theta=\frac{\pi}{6},\frac{5\pi}{6} \).
3. Sum \( =0+\frac{\pi}{6}+\frac{\pi}{3}+\frac{2\pi}{3}+\frac{5\pi}{6}+\pi=3\pi \).

Teacher's Note:
1. List the solutions of each factor inside the given interval and check for repeats before adding.
2. There are six distinct solutions here.

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

Explanation:
1. \( \cos2x-1=-2\sin^2x \) and \( \cot^2x=\frac{\cos^2x}{\sin^2x} \), so the left side is \( -2\cos^2x \) (for \( \sin x\ne0 \)).
2. \( -2\cos^2x=-3\sin x\Rightarrow2\sin^2x+3\sin x-2=0\Rightarrow(2\sin x-1)(\sin x+2)=0 \).
3. \( \sin x=-2 \) is impossible, so \( \sin x=\frac12\Rightarrow x=n\pi+(-1)^n\frac{\pi}{6} \).

Teacher's Note:
1. Simplify the left side before solving.
2. The root sin x = −2 lies outside [−1, 1] and is rejected, and sin x = 0 is excluded by cot x.