JEE Mathematics Trigonometric Equations MCQs Set 08

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

Explanation:
1. \( \sec^2\theta=\frac43\Rightarrow\cos^2\theta=\frac34=\cos^2\frac{\pi}{6} \).
2. Use \( \cos^2\theta=\cos^2\alpha\Rightarrow\theta=n\pi\pm\alpha \).
3. \( \theta=n\pi\pm\frac{\pi}{6} \).

Teacher's Note:
1. For a squared trigonometric function the general solution is always nπ ± α.
2. Do not use 2nπ here, since the sign of the cosine is not fixed.

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

Explanation:
1. Move all terms to one side: \( (1-\sin x)-\cos2x(1-\sin x)=0 \), i.e. \( (1-\sin x)(1-\cos2x)=0 \).
2. \( \sin x=1\Rightarrow x=2n\pi+\frac{\pi}{2} \).
3. \( \cos2x=1\Rightarrow2x=2n\pi\Rightarrow x=n\pi \).

Teacher's Note:
1. Group terms so that (1 − sin x) appears twice.
2. The two factors give a family each, and they do not overlap.

Answer: (a) \( 30^\circ,45^\circ,90^\circ,135^\circ,150^\circ \)

Explanation:
1. \( \cos6\theta+\cos2\theta=2\cos4\theta\cos2\theta \) and \( \cos4\theta+1=2\cos^22\theta \), so the left side is \( 2\cos2\theta(\cos4\theta+\cos2\theta)=4\cos2\theta\cos3\theta\cos\theta \).
2. \( \cos2\theta=0\Rightarrow\theta=45^\circ,135^\circ \); \( \cos3\theta=0\Rightarrow\theta=30^\circ,90^\circ,150^\circ \); \( \cos\theta=0\Rightarrow\theta=90^\circ \).
3. Together: \( 30^\circ,45^\circ,90^\circ,135^\circ,150^\circ \).

Teacher's Note:
1. Factorise into a product of three cosines and list the zeros of each in the given interval. θ = 90° appears twice but is counted once.

Answer: (d) \( x = 2n\pi \) or \( x = \frac{2n\pi}{3}-\frac{\pi}{6} \)

Explanation:
1. \( (\sin2\theta-\sin\theta)-(\cos2\theta-\cos\theta)=2\cos\frac{3\theta}{2}\sin\frac{\theta}{2}+2\sin\frac{3\theta}{2}\sin\frac{\theta}{2} \).
2. \( =2\sin\frac{\theta}{2}\left[\cos\frac{3\theta}{2}+\sin\frac{3\theta}{2}\right]=0 \).
3. \( \sin\frac{\theta}{2}=0\Rightarrow\theta=2n\pi \); \( \tan\frac{3\theta}{2}=-1\Rightarrow\frac{3\theta}{2}=n\pi-\frac{\pi}{4}\Rightarrow\theta=\frac{2n\pi}{3}-\frac{\pi}{6} \).

Teacher's Note:
1. Group the sine terms together and the cosine terms together before using the difference formulas.
2. The common half-angle factor sin(θ/2) appears.

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

Explanation:
1. Use \( 2\sin A\sin B=\cos(A-B)-\cos(A+B) \): \( \cos6x-\cos8x=\cos2x-\cos8x \).
2. \( \cos6x=\cos2x\Rightarrow6x=2n\pi\pm2x \).
3. \( x=\frac{n\pi}{2} \) or \( x=\frac{n\pi}{4} \); the second family contains the first, so \( x=\frac{n\pi}{4} \).

Teacher's Note:
1. Both sides turn into a difference of cosines with the same cos 8x, which cancels.
2. The larger family nπ/4 includes nπ/2.

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

Explanation:
1. \( \sin4\theta-\sin2\theta=2\cos3\theta\sin\theta \), so \( \cos3\theta(2\sin\theta-1)=0 \).
2. \( \cos3\theta=0\Rightarrow\theta=\frac{(2n+1)\pi}{6} \).
3. \( \sin\theta=\frac12\Rightarrow\theta=n\pi+(-1)^n\frac{\pi}{6} \).

Teacher's Note:
1. The right side is the same cos 3θ that appears from the difference formula, so it factors out.
2. Note that θ = π/6 belongs to both families.

Answer: (a) \( x = 2n\pi\pm\frac{\pi}{6} \) or \( x = 2n\pi\pm\frac{\pi}{3} \)

Explanation:
1. Factorise: \( 4\cos^2\theta-2(\sqrt3+1)\cos\theta+\sqrt3=(2\cos\theta-\sqrt3)(2\cos\theta-1)=0 \).
2. \( \cos\theta=\frac{\sqrt3}{2}\Rightarrow\theta=2n\pi\pm\frac{\pi}{6} \).
3. \( \cos\theta=\frac12\Rightarrow\theta=2n\pi\pm\frac{\pi}{3} \).

Teacher's Note:
1. Check by expansion: (2c − √3)(2c − 1) = 4c² − 2(√3 + 1)c + √3.
2. Each cosine value gives its own family.

Answer: (c) \( x = \frac{n\pi}{2}\pm\frac{\pi}{6} \)

Explanation:
1. With \( y=\sin^2x\cos^2x=\frac{\sin^22x}{4} \): \( f=(1-0.7)-(3-1.4)y=0.3-1.6y=0 \).
2. \( y=\frac{3}{16}\Rightarrow\sin^22x=\frac34 \).
3. \( 2x=n\pi\pm\frac{\pi}{3}\Rightarrow x=\frac{n\pi}{2}\pm\frac{\pi}{6} \).

Teacher's Note:
1. Substituting y = sin²x cos²x reduces both fourth and sixth powers to polynomials in y.
2. The condition sin²2x = 3/4 then gives 2x = nπ ± π/3.

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

Explanation:
1. \( \cot^2\theta-\tan^2\theta=(\cot\theta-\tan\theta)(\cot\theta+\tan\theta)=2\cot2\theta\cdot\frac{2}{\sin2\theta} \).
2. So \( 4\cot2\theta\left(1-\frac{1}{\sin2\theta}\right)=0 \): \( \cot2\theta=0 \) or \( \sin2\theta=1 \).
3. \( \cot2\theta=0\Rightarrow\theta=\frac{(2n+1)\pi}{4} \); \( \sin2\theta=1\Rightarrow\theta=n\pi+\frac{\pi}{4} \), which is contained in the first family.

Teacher's Note:
1. Use cot θ + tan θ = 2/sin 2θ and cot θ − tan θ = 2 cot 2θ.
2. The second family is a subset of the first.

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

Explanation:
1. \( \cot\theta+\tan\theta=\frac{\cos^2\theta+\sin^2\theta}{\sin\theta\cos\theta}=\frac{1}{\sin\theta\cos\theta} \).
2. \( \frac{1}{\sin\theta\cos\theta}=\frac{2}{\sin\theta}\Rightarrow\cos\theta=\frac12 \) (with \( \sin\theta\ne0 \)).
3. \( \theta=2n\pi\pm\frac{\pi}{3} \).

Teacher's Note:
1. cot θ + tan θ simplifies to 1/(sin θ cos θ), which makes cancellation with cosec θ easy.
2. Since cos θ = ½ fixes the sign, the period is 2π.

Answer: (b) \( x = n\pi \) or \( x = \frac{(2n+1)\pi}{20} \)

Explanation:
1. With \( \cos5x\ne0 \): \( \sin5x\cos3x=\sin7x\cos5x\Rightarrow\sin8x+\sin2x=\sin12x+\sin2x \).
2. \( \sin12x=\sin8x\Rightarrow4x=2m\pi\text{ or }20x=(2m+1)\pi \), i.e. \( x=\frac{m\pi}{2} \) or \( x=\frac{(2m+1)\pi}{20} \).
3. For odd m, \( x=\frac{m\pi}{2} \) makes \( \cos5x=0 \) and must be rejected, leaving \( x=n\pi \). The second family never makes \( \cos5x \) zero.

Teacher's Note:
1. Clearing the denominator of tan 5x creates extra roots where cos 5x = 0.
2. Always test them against the original equation.

Answer: (c) \( x = 2n\pi+\frac{\pi}{4} \) or \( x = n\pi+(-1)^{n+1}\frac{\pi}{6}-\frac{\pi}{4} \)

Explanation:
1. Multiply by \( \sin x\cos x \) and put \( u=\sin x+\cos x \), \( \sin x\cos x=\frac{u^2-1}{2} \): \( u=\sqrt2(u^2-1) \).
2. \( \sqrt2u^2-u-\sqrt2=0\Rightarrow u=\sqrt2 \) or \( u=-\frac{1}{\sqrt2} \).
3. \( u=\sqrt2\Rightarrow\sin\left(x+\frac{\pi}{4}\right)=1\Rightarrow x=2n\pi+\frac{\pi}{4} \); \( u=-\frac{1}{\sqrt2}\Rightarrow\sin\left(x+\frac{\pi}{4}\right)=-\frac12\Rightarrow x=n\pi+(-1)^{n+1}\frac{\pi}{6}-\frac{\pi}{4} \).

Teacher's Note:
1. Equations in sin x + cos x and sin x cos x are best handled by the substitution u = sin x + cos x.
2. Both roots of the resulting quadratic are valid here.

Answer: (c) \( x = n\pi-\frac{\pi}{4} \) or \( x = n\pi\pm\frac{\pi}{3} \)

Explanation:
1. Use \( \cot(\frac{\pi}{2}-x)=\tan x \) and \( \sec^2x=1+\tan^2x \), with \( t=\tan x \): \( t^3-1+1+t^2-3t-3=0 \).
2. \( t^3+t^2-3t-3=(t+1)(t^2-3)=0 \).
3. \( t=-1\Rightarrow x=n\pi-\frac{\pi}{4} \); \( t=\pm\sqrt3\Rightarrow x=n\pi\pm\frac{\pi}{3} \).

Teacher's Note:
1. Reduce everything to tan x and factorise the cubic by grouping.
2. Remember that tan x = ±√3 gives x = nπ ± π/3.

Answer: (c) \( x = 2n\pi+\frac{3\pi}{4} \)

Explanation:
1. \( \cos\left(x+\frac{3\pi}{4}\right)=-\cos\left(x-\frac{\pi}{4}\right) \), so the equation is \( \sin\left(x-\frac{\pi}{4}\right)+\cos\left(x-\frac{\pi}{4}\right)=\sqrt2\sin x=1 \), giving \( x=n\pi+(-1)^n\frac{\pi}{4} \).
2. At these values \( \cos2x=0 \), so the right side of the inequality is \( 1 \). Also \( \cos3+\sin3\approx-0.85\lt0 \), so the inequality needs \( \cos7x\lt\frac{\cos3+\sin3}{2}\approx-0.42 \).
3. \( x=\frac{\pi}{4}+2k\pi \): \( \cos7x=\cos\frac{7\pi}{4}=0.71 \) ✗. \( x=\frac{3\pi}{4}+2k\pi \): \( \cos7x=\cos\frac{21\pi}{4}=-0.71 \) ✓. So \( x=2n\pi+\frac{3\pi}{4} \).

Teacher's Note:
1. Solve the equality first, then use the inequality to select among the solutions.
2. Note that cos 3 + sin 3 is negative (3 radians is close to π), which reverses the inequality.

Answer: (b) \( 2n\pi\pm\frac{\pi}{6} \)

Explanation:
1. For even \( n=2k \): \( (4k-1)\frac{\pi}{2}+\frac{\pi}{3}=2k\pi-\frac{\pi}{2}+\frac{\pi}{3}=2k\pi-\frac{\pi}{6} \).
2. For odd \( n=2k+1 \): \( (4k+1)\frac{\pi}{2}-\frac{\pi}{3}=2k\pi+\frac{\pi}{2}-\frac{\pi}{3}=2k\pi+\frac{\pi}{6} \).
3. Together these are \( 2k\pi\pm\frac{\pi}{6} \).

Teacher's Note:
1. Split a formula with (−1)ⁿ into the even and odd cases of n.
2. Each case reduces to a simple multiple of 2π plus a fixed angle.

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

Explanation:
1. \( \cot\theta-\tan\theta=2\cot2\theta \).
2. \( 2\cot2\theta=2\Rightarrow\cot2\theta=1\Rightarrow2\theta=n\pi+\frac{\pi}{4} \).
3. \( \theta=\frac{n\pi}{2}+\frac{\pi}{8} \).

Teacher's Note:
1. The identity cot θ − tan θ = 2 cot 2θ turns this into a basic equation.
2. Remember to halve both nπ and π/4.

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

Explanation:
1. Factorise: \( \sqrt3(\sin x-1)+\tan x(\sin x-1)=(\sin x-1)(\sqrt3+\tan x)=0 \).
2. \( \sin x=1 \) gives \( x=\frac{\pi}{2} \), where \( \tan x \) is undefined, so it must be rejected.
3. \( \tan x=-\sqrt3\Rightarrow x=n\pi-\frac{\pi}{3} \).

Teacher's Note:
1. Always check every candidate root against the domain of the original equation.
2. Here tan x is undefined at the root of sin x = 1.

Answer: (b) \( x = \frac{n\pi}{4} \) or \( x = \frac{2n\pi}{3}\pm\frac{\pi}{9} \)

Explanation:
1. \( \sin7\theta+\sin\theta=2\sin4\theta\cos3\theta \), so \( \sin4\theta(2\cos3\theta-1)=0 \).
2. \( \sin4\theta=0\Rightarrow\theta=\frac{n\pi}{4} \).
3. \( \cos3\theta=\frac12\Rightarrow3\theta=2n\pi\pm\frac{\pi}{3}\Rightarrow\theta=\frac{2n\pi}{3}\pm\frac{\pi}{9} \).

Teacher's Note:
1. Combine sin 7θ and sin θ so that sin 4θ becomes a common factor.
2. Divide the whole second general solution by 3.

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

Explanation:
1. For the logarithms to exist, \( \sin x\gt0 \) and \( \cos x\gt0 \). Then \( 5^{\log_5\sin x}=\sin x \) and \( 15^{\log_{15}\cos x}=\cos x \).
2. \( \sqrt5+\sqrt5\sin x=\sqrt{15}\cos x\Rightarrow1+\sin x=\sqrt3\cos x\Rightarrow\sqrt3\cos x-\sin x=1 \), i.e. \( \cos\left(x+\frac{\pi}{6}\right)=\frac12 \).
3. \( x=2n\pi+\frac{\pi}{6} \) or \( x=2n\pi-\frac{\pi}{2} \). The second has \( \sin x=-1\lt0 \) and is rejected. So \( x=2n\pi+\frac{\pi}{6} \).

Teacher's Note:
1. Logarithms need positive arguments, so the domain sin x > 0, cos x > 0 must be applied at the end.
2. The root x = −π/2 fails it.

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

Explanation:
1. Equal sets have equal sums: \( (\sin\alpha+\sin3\alpha+\sin2\alpha)=(\cos\alpha+\cos3\alpha+\cos2\alpha)\Rightarrow\sin2\alpha(2\cos\alpha+1)=\cos2\alpha(2\cos\alpha+1) \).
2. So \( \cos\alpha=-\frac12 \) or \( \tan2\alpha=1 \). For \( \cos\alpha=-\frac12 \), e.g. \( \alpha=\frac{2\pi}{3} \): the sine set contains \( \frac{\sqrt3}{2} \) but the cosine set is \( \left\{-\frac12,-\frac12,1\right\} \): not equal. So reject.
3. \( \tan2\alpha=1\Rightarrow\alpha=\frac{n\pi}{2}+\frac{\pi}{8} \): for \( \alpha=\frac{\pi}{8} \) both sets are \( \{\sin22.5^\circ,\sin45^\circ,\sin67.5^\circ\}=\{\cos67.5^\circ,\cos45^\circ,\cos22.5^\circ\} \). The same check works for the other three values in a period.

Teacher's Note:
1. Equality of sets forces equality of sums, which gives a necessary condition.
2. Then each candidate must be tested for the full set.

Answer: (a) 10

Explanation:
1. \( \sin x+\sin5x-\sin3x=\sin3x(2\cos2x-1) \) and \( \cos x+\cos5x-\cos3x=\cos3x(2\cos2x-1) \).
2. So \( (2\cos2x-1)(\sin3x-\cos3x)=0 \).
3. \( \cos2x=\frac12\Rightarrow x=\frac{\pi}{6},\frac{5\pi}{6},\frac{7\pi}{6},\frac{11\pi}{6} \) (4 values). \( \tan3x=1\Rightarrow x=\frac{\pi}{12}+\frac{k\pi}{3} \), \( k=0,\ldots,5 \) (6 values). The two sets are disjoint, so the total is 10.

Teacher's Note:
1. Count each family separately within one period and make sure no value is counted twice. tan 3x = 1 gives six values in [0, 2π).

Answer: (c) \( \left[-\frac32,\frac12\right] \)

Explanation:
1. \( \sin^4x+\cos^4x=1-\frac{\sin^22x}{2} \). With \( u=\sin2x\in[-1,1] \): \( a=\frac{u^2}{2}-u-1 \).
2. \( g(u)=\frac{u^2}{2}-u-1 \) has \( g'(u)=u-1\le0 \) on \( [-1,1] \), so it decreases from \( g(-1)=\frac12 \) to \( g(1)=-\frac32 \).
3. So \( a\in\left[-\frac32,\frac12\right] \). (For such \( a \): \( \sin2x=1-\sqrt{3+2a} \).)

Teacher's Note:
1. Write the equation as a quadratic in sin 2x and find the range of the resulting expression for sin 2x in [−1, 1].
2. A decreasing function takes its extreme values at the endpoints.

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

Explanation:
1. \( \left(\tan\theta-\cot\theta\right)^2=\tan^2\theta+\cot^2\theta-2=0 \).
2. \( \tan\theta=\cot\theta\Rightarrow\tan^2\theta=1 \).
3. \( \theta=n\pi\pm\frac{\pi}{4}=\frac{n\pi}{2}+\frac{\pi}{4} \).

Teacher's Note:
1. The left side minus 2 is a perfect square, so tan θ = cot θ.
2. The two families nπ + π/4 and nπ − π/4 together make up nπ/2 + π/4.

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

Explanation:
1. A quadratic \( ax^2+bx+c \) with \( a\ne0 \) is a perfect square when \( b^2=4ac \): \( 4\cos^2\theta=4\sin\theta\cdot\frac{\cos\theta+\sin\theta}{2} \).
2. \( 2\cos^2\theta-\sin\theta\cos\theta-\sin^2\theta=0\Rightarrow(2\cos\theta+\sin\theta)(\cos\theta-\sin\theta)=0 \).
3. \( \tan\theta=1\Rightarrow\theta=n\pi+\frac{\pi}{4} \); \( \tan\theta=-2\Rightarrow\theta=n\pi-\tan^{-1}2 \). Here \( \sin\theta\ne0 \) in both cases, so the quadratic is a true quadratic.

Teacher's Note:
1. A perfect square quadratic has discriminant zero.
2. Divide by cos²θ after factorising to get the two values of tan θ.

Answer: (a) the same set of angles

Explanation:
1. Put \( n=2k \) in the second formula: \( 2k\pi+\frac{\pi}{2}-\alpha \). Put \( n=2k+1 \): \( (2k+1)\pi-\frac{\pi}{2}+\alpha=2k\pi+\frac{\pi}{2}+\alpha \).
2. Together: \( 2k\pi+\frac{\pi}{2}\pm\alpha=\left(2k+\frac12\right)\pi\pm\alpha \), which is the first formula.
3. So both formulae give the same angles for every \( \alpha \).

Teacher's Note:
1. Both formulae describe all solutions of sin θ = cos α.
2. Splitting n into even and odd values proves they coincide.

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

Explanation:
1. Multiply by \( \sin x \): \( 1=\sin x+\cos x\Rightarrow\sin\left(x+\frac{\pi}{4}\right)=\frac{1}{\sqrt2} \).
2. \( x+\frac{\pi}{4}=n\pi+(-1)^n\frac{\pi}{4} \): even \( n \) gives \( x=2m\pi \), odd \( n \) gives \( x=2m\pi+\frac{\pi}{2} \).
3. \( x=2m\pi \) makes \( \csc x \) undefined, so \( x=2n\pi+\frac{\pi}{2} \).

Teacher's Note:
1. The same trap as in the previous problem: the root x = 2nπ comes from clearing the denominator and is not a root of the original equation.