JEE Mathematics Trigonometric Equations MCQs Set 07

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

Explanation:
1. Numerator: \( (1+\cos2x)+\sin x+\cos x+2\sin x\cos x=2\cos^2x+2\sin x\cos x+\sin x+\cos x=(\sin x+\cos x)(2\cos x+1) \).
2. \( \tan x=-1 \) would make \( 2x=-\frac{\pi}{2}+2n\pi \) and \( \tan2x \) undefined, so it is rejected.
3. \( \cos x=-\frac12\Rightarrow x=2n\pi\pm\frac{2\pi}{3} \), where \( \tan2x=\pm\sqrt3 \) is defined and non-zero.

Teacher's Note:
1. A fraction is zero only when the numerator is zero and the denominator is defined and non-zero.
2. Test each family against tan 2x.

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

Explanation:
1. \( 4\cos^3\theta-3\cos\theta+2\cos\theta=4\cos^3\theta-\cos\theta=\cos\theta(2\cos\theta-1)(2\cos\theta+1)=0 \).
2. \( \cos\theta=0\Rightarrow\theta=\frac{(2n+1)\pi}{2} \).
3. \( \cos\theta=\pm\frac12\Rightarrow\cos^2\theta=\frac14\Rightarrow\theta=n\pi\pm\frac{\pi}{3} \).

Teacher's Note:
1. Use cos 3θ = 4cos³θ − 3cos θ and factorise.
2. The two values cos θ = ±½ are combined into cos²θ = ¼.

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

Explanation:
1. \( 2\cos3x\cos x=\cos4x+\cos2x \).
2. The equation becomes \( 1+\cos4x=0 \), so \( \cos4x=-1 \).
3. \( 4x=(2n+1)\pi\Rightarrow x=\frac{(2n+1)\pi}{4} \).

Teacher's Note:
1. The cos 2x terms cancel after the product is converted into a sum.
2. What remains is a single standard equation.

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

Explanation:
1. Factorise: \( (\cot\theta+\sqrt3)\left(\cot\theta+\frac{1}{\sqrt3}\right)=0 \) (product of roots \( \sqrt3\cdot\frac{1}{\sqrt3}=1 \)).
2. \( \cot\theta=-\sqrt3\Rightarrow\tan\theta=-\frac{1}{\sqrt3}\Rightarrow\theta=n\pi-\frac{\pi}{6} \).
3. \( \cot\theta=-\frac{1}{\sqrt3}\Rightarrow\tan\theta=-\sqrt3\Rightarrow\theta=n\pi-\frac{\pi}{3} \).

Teacher's Note:
1. Look for two numbers with sum √3 + 1/√3 and product 1 before using the quadratic formula.
2. Convert cot to tan to read off the angle.

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

Explanation:
1. Base \( b=\frac{-x^2-6x}{10} \) must satisfy \( b\gt0 \), \( b\ne1 \): \( -6\lt x\lt0 \); and \( b=1 \) would need \( x^2+6x+10=0 \), which has no real root.
2. \( \sin3x+\sin x=\sin2x\Rightarrow\sin2x(2\cos x-1)=0 \), with \( \sin2x\gt0 \) required for the logarithm. So \( \cos x=\frac12 \), \( x=2n\pi\pm\frac{\pi}{3} \).
3. In \( (-6,0) \) the candidates are \( -\frac{\pi}{3} \) (then \( \sin2x=-\frac{\sqrt3}{2}\lt0 \), rejected) and \( -\frac{5\pi}{3}\approx-5.24 \) (then \( \sin\left(-\frac{10\pi}{3}\right)=\frac{\sqrt3}{2}\gt0 \), accepted). \( -\frac{7\pi}{3}\approx-7.33 \) is outside the interval.

Teacher's Note:
1. Logarithm equations need the base conditions and positive arguments before any trigonometry is applied.
2. Check the sign of sin 2x for each candidate.

Answer: (d) \( x = n\pi+\frac{\pi}{12} \) or \( x = n\pi+\frac{5\pi}{12} \)

Explanation:
1. With \( t=\tan x \): \( t+\frac{1+t}{1-t}=2\Rightarrow t(1-t)+1+t=2(1-t) \).
2. \( t^2-4t+1=0\Rightarrow t=2\pm\sqrt3 \).
3. \( \tan\frac{5\pi}{12}=2+\sqrt3 \) and \( \tan\frac{\pi}{12}=2-\sqrt3 \), so \( x=n\pi+\frac{\pi}{12} \) or \( x=n\pi+\frac{5\pi}{12} \).

Teacher's Note:
1. The fraction is just tan(π/4 + x).
2. Recall tan 15° = 2 − √3 and tan 75° = 2 + √3.

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

Explanation:
1. \( \cos\theta-\cos2\theta=2\sin\frac{3\theta}{2}\sin\frac{\theta}{2} \) and \( \sin3\theta=2\sin\frac{3\theta}{2}\cos\frac{3\theta}{2} \), so \( \sin\frac{3\theta}{2}\left[\sin\frac{\theta}{2}-\cos\frac{3\theta}{2}\right]=0 \).
2. \( \sin\frac{3\theta}{2}=0\Rightarrow\theta=\frac{2n\pi}{3} \).
3. \( \cos\frac{3\theta}{2}=\cos\left(\frac{\pi}{2}-\frac{\theta}{2}\right)\Rightarrow\frac{3\theta}{2}=2n\pi\pm\left(\frac{\pi}{2}-\frac{\theta}{2}\right) \): \( \theta=n\pi+\frac{\pi}{4} \) or \( \theta=2n\pi-\frac{\pi}{2} \).

Teacher's Note:
1. Use half-angle products on both sides to expose sin(3θ/2).
2. Handle the last factor by writing sin(θ/2) as a cosine.

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

Explanation:
1. \( \sin4x-\sin2x=2\cos3x\sin x \), so \( \cos3x-2\cos3x\sin x=\cos3x(1-2\sin x)=0 \).
2. \( \cos3x=0\Rightarrow3x=(2n+1)\frac{\pi}{2}\Rightarrow x=\frac{(2n+1)\pi}{6} \).
3. \( \sin x=\frac12\Rightarrow x=n\pi+(-1)^n\frac{\pi}{6} \).

Teacher's Note:
1. The difference sin 4x − sin 2x produces cos 3x, which is already present in the equation.
2. That makes the factorisation immediate.

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

Explanation:
1. \( 2\sqrt3(1-\sin^2\theta)=\sin\theta\Rightarrow2\sqrt3\sin^2\theta+\sin\theta-2\sqrt3=0 \).
2. \( \sin\theta=\frac{-1\pm7}{4\sqrt3} \), i.e. \( \frac{\sqrt3}{2} \) or \( -\frac{2}{\sqrt3} \) (impossible since \( \frac{2}{\sqrt3}\gt1 \)).
3. \( \sin\theta=\frac{\sqrt3}{2}\Rightarrow\theta=n\pi+(-1)^n\frac{\pi}{3} \).

Teacher's Note:
1. Use the quadratic formula when the factors are not obvious.
2. The discriminant here is 49, a perfect square, which confirms the roots are neat.

Answer: (b) \( 45^\circ \) and \( 60^\circ \)

Explanation:
1. Multiply by \( \sin^2\theta\cos^2\theta \): \( 1+2\cos^2\theta=8\sin^2\theta\cos^2\theta \). With \( u=\cos^2\theta \): \( 1+2u=8u(1-u)\Rightarrow8u^2-6u+1=0 \).
2. \( (2u-1)(4u-1)=0\Rightarrow u=\frac12 \) or \( u=\frac14 \).
3. \( \cos^2\theta=\frac12\Rightarrow\theta=45^\circ \); \( \cos^2\theta=\frac14\Rightarrow\theta=60^\circ \).

Teacher's Note:
1. Clear the denominators to get a quadratic in cos²θ.
2. Both roots give an acute angle, so both are accepted.

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

Explanation:
1. \( \cos3\theta+\cos\theta=2\cos2\theta\cos\theta \), so \( 2\cos2\theta(\cos\theta-1)=0 \).
2. \( \cos2\theta=0\Rightarrow\theta=\frac{(2n+1)\pi}{4} \).
3. \( \cos\theta=1\Rightarrow\theta=2n\pi \).

Teacher's Note:
1. Do not divide by cos 2θ; factorise instead.
2. The family 2nπ is not contained in the odd multiples of π/4.

Answer: (a) \( x = n\pi-\tan^{-1}\frac12 \)

Explanation:
1. Collect terms: \( \cos^2x+2\sin x\cos x-2\cos x-4\sin x=0 \).
2. Factorise: \( \cos x(\cos x+2\sin x)-2(\cos x+2\sin x)=(\cos x-2)(\cos x+2\sin x)=0 \).
3. \( \cos x=2 \) is impossible; \( \tan x=-\frac12\Rightarrow x=n\pi-\tan^{-1}\frac12 \).

Teacher's Note:
1. Writing sin 2x = 2 sin x cos x lets the left side factorise by grouping.
2. One factor can never vanish, since cos x cannot equal 2.

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

Explanation:
1. Let \( y=\sin^2x\cos^2x\in\left[0,\frac14\right] \). Then \( \sin^6x+\cos^6x=1-3y \) and \( \sin^4x+\cos^4x=1-2y \), so \( f=(1+k)-(3+2k)y \).
2. This is linear in \( y \), so its values lie between \( f(0)=1+k \) and \( f\left(\frac14\right)=\frac{1+2k}{4} \). Zero is attained iff these two have opposite signs (or one is zero): \( (1+k)(1+2k)\le0 \).
3. \( -1\le k\le-\frac12 \).

Teacher's Note:
1. Reduce f to a linear function of y = sin²x cos²x and use the endpoints of its range.
2. The product of the endpoint values must be non-positive.

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

Explanation:
1. \( 2\frac{\cos^2\theta}{\sin^2\theta}=\frac{1}{\sin^2\theta}\Rightarrow2\cos^2\theta=1 \) (for \( \sin\theta\ne0 \)).
2. \( \cos2\theta=2\cos^2\theta-1=0\Rightarrow2\theta=(2n+1)\frac{\pi}{2} \).
3. \( \theta=\frac{n\pi}{2}+\frac{\pi}{4} \).

Teacher's Note:
1. Convert both sides to sine and cosine and cancel the common sin²θ.
2. The equation 2cos²θ = 1 is the same as cos 2θ = 0.

Answer: (c) \( x = \frac{(2n+1)\pi}{4} \) or \( x = 2n\pi\pm\frac{2\pi}{3} \)

Explanation:
1. \( \cos3\theta+\cos\theta=2\cos2\theta\cos\theta \), so \( \cos2\theta(2\cos\theta+1)=0 \).
2. \( \cos2\theta=0\Rightarrow\theta=\frac{(2n+1)\pi}{4} \).
3. \( \cos\theta=-\frac12\Rightarrow\theta=2n\pi\pm\frac{2\pi}{3} \).

Teacher's Note:
1. Pair the outer terms, then factor out the middle term cos 2θ.
2. The sum of three cosines in AP always factors this way.

Answer: (a) \( x = \frac{(2n+1)\pi}{14} \) or \( x = \frac{2n\pi}{7}\pm\frac{\pi}{28} \)

Explanation:
1. Factorise: \( \cos7x(\sqrt2\cos7x-1)=0 \).
2. \( \cos7x=0\Rightarrow7x=(2n+1)\frac{\pi}{2}\Rightarrow x=\frac{(2n+1)\pi}{14} \).
3. \( \cos7x=\frac{1}{\sqrt2}\Rightarrow7x=2n\pi\pm\frac{\pi}{4}\Rightarrow x=\frac{2n\pi}{7}\pm\frac{\pi}{28} \).

Teacher's Note:
1. Never divide by cos 7x; factorise so that no family of roots is lost.
2. Each factor then gives its own family.

Answer: (a) \( x = \frac{(4n+3)\pi}{10} \)

Explanation:
1. \( \cos2\theta=\sin\left(\frac{\pi}{2}-2\theta\right) \), so \( \sin3\theta=-\sin\left(\frac{\pi}{2}-2\theta\right)=\sin\left(2\theta-\frac{\pi}{2}\right) \).
2. \( 3\theta=n\pi+(-1)^n\left(2\theta-\frac{\pi}{2}\right) \). Even \( n=2m \): \( \theta=2m\pi-\frac{\pi}{2} \). Odd \( n=2m+1 \): \( 5\theta=(2m+1)\pi+\frac{\pi}{2}\Rightarrow\theta=\frac{(4m+3)\pi}{10} \).
3. The even-case family is already contained in the odd-case family: \( 2m\pi-\frac{\pi}{2}=\frac{(4k+3)\pi}{10} \) with \( k=5m-2 \). So \( \theta=\frac{(4n+3)\pi}{10} \).

Teacher's Note:
1. Rewrite the cosine as a sine and use sin A = sin B.
2. Always check whether one family lies inside another before listing both.

Answer: (a) \( x = \frac{(2n+1)\pi}{8} \) or \( x = n\pi\pm\frac{\pi}{3} \)

Explanation:
1. \( \frac{3-\cos2x-\cos4x-\cos6x}{2}=\frac32\Rightarrow\cos2x+\cos4x+\cos6x=0 \).
2. \( 2\cos4x\cos2x+\cos4x=\cos4x(2\cos2x+1)=0 \).
3. \( \cos4x=0\Rightarrow x=\frac{(2n+1)\pi}{8} \); \( \cos2x=-\frac12\Rightarrow2x=2n\pi\pm\frac{2\pi}{3}\Rightarrow x=n\pi\pm\frac{\pi}{3} \).

Teacher's Note:
1. Convert squares to double angles, then group cos 6x with cos 2x.
2. Remember that cos θ = −½ gives θ = 2nπ ± 2π/3.

Answer: (a) \( x = 2n\pi\pm\cos^{-1}\frac{\sqrt{17}-3}{4} \)

Explanation:
1. \( 2\cos^2\theta-1+3\cos\theta=0\Rightarrow2\cos^2\theta+3\cos\theta-1=0 \).
2. \( \cos\theta=\frac{-3\pm\sqrt{17}}{4} \); the value \( \frac{-3-\sqrt{17}}{4}\approx-1.78 \) is rejected.
3. \( \cos\theta=\frac{\sqrt{17}-3}{4}\Rightarrow\theta=2n\pi\pm\cos^{-1}\frac{\sqrt{17}-3}{4} \).

Teacher's Note:
1. The discriminant 9 + 8 = 17 is not a perfect square, so the answer is left in inverse-cosine form.
2. Always test the second root against −1.

Answer: (d) \( x = \frac{(4n+1)\pi}{10} \)

Explanation:
1. \( \cos2x=\sin3x=\cos\left(\frac{\pi}{2}-3x\right) \).
2. \( 2x=2n\pi\pm\left(\frac{\pi}{2}-3x\right) \). With the plus sign, \( 5x=2n\pi+\frac{\pi}{2}\Rightarrow x=\frac{(4n+1)\pi}{10} \).
3. With the minus sign, \( x=\frac{\pi}{2}-2n\pi \) ; these values are already of the form \( \frac{(4k+1)\pi}{10} \) (take \( k=5m+1 \)). So \( x=\frac{(4n+1)\pi}{10} \).

Teacher's Note:
1. Write sin 3x as cos(π/2 − 3x) and use cos A = cos B.
2. Check whether the second family is already contained in the first before listing it.

Answer: (d) \( x = \frac{(2n+1)\pi}{8} \) or \( x = \frac{n\pi}{3} \)

Explanation:
1. \( \cos6x+\cos2x=2\cos4x\cos2x \) and \( \cos8x+1=2\cos^24x \).
2. So \( 2\cos4x(\cos2x-\cos4x)=0 \).
3. \( \cos4x=0\Rightarrow x=\frac{(2n+1)\pi}{8} \); \( \cos2x=\cos4x\Rightarrow4x=2n\pi\pm2x\Rightarrow x=n\pi\text{ or }\frac{n\pi}{3} \), and \( \frac{n\pi}{3} \) already contains \( n\pi \).

Teacher's Note:
1. Pair cos 6x with cos 2x and use 1 + cos 8x = 2cos²4x so that cos 4x is common.
2. When two families overlap, state the larger one.

Answer: (c) \( x = \frac{n\pi}{2}+\frac{\pi}{8} \) or \( x = \frac{n\pi}{4}+(-1)^{n+1}\frac{\pi}{24} \)

Explanation:
1. Let \( u=\tan2x+\cot2x \): \( u^2-2+2u=6\Rightarrow u^2+2u-8=0\Rightarrow(u-2)(u+4)=0 \).
2. \( u=\tan2x+\cot2x=\frac{2}{\sin4x} \), so \( \sin4x=1 \) or \( \sin4x=-\frac12 \).
3. \( \sin4x=1\Rightarrow x=\frac{n\pi}{2}+\frac{\pi}{8} \); \( \sin4x=-\frac12\Rightarrow4x=n\pi+(-1)^{n+1}\frac{\pi}{6}\Rightarrow x=\frac{n\pi}{4}+(-1)^{n+1}\frac{\pi}{24} \).

Teacher's Note:
1. Group tan 2x + cot 2x as u, noting that the square of u gives tan² + cot² + 2.
2. Then use u = 2/ sin 4x.

Answer: (d) \( x = 2n\pi+\frac{\pi}{2} \) or \( x = n\pi-(-1)^n\sin^{-1}\frac13 \)

Explanation:
1. \( 3\sin^2\theta-2\sin\theta-1=(3\sin\theta+1)(\sin\theta-1)=0 \).
2. \( \sin\theta=1\Rightarrow\theta=2n\pi+\frac{\pi}{2} \).
3. \( \sin\theta=-\frac13=\sin\left(-\sin^{-1}\frac13\right)\Rightarrow\theta=n\pi+(-1)^n\left(-\sin^{-1}\frac13\right)=n\pi-(-1)^n\sin^{-1}\frac13 \).

Teacher's Note:
1. For sin θ = −k, the principal value is −sin⁻¹k, which turns (−1)ⁿ into a minus sign in front of it.

Answer: (d) \( x = \frac{n\pi}{3}+(-1)^{n+1}\frac{\pi}{18} \)

Explanation:
1. Let \( u=\sin3x \). Squaring: \( 16u^2+24u+9=2u+2\Rightarrow16u^2+22u+7=0\Rightarrow(2u+1)(8u+7)=0 \).
2. So \( u=-\frac12 \) or \( u=-\frac78 \). The left side \( 4u+3 \) must be non-negative: for \( u=-\frac78 \) it is \( -\frac12 \), so this root is extraneous.
3. \( \sin3x=-\frac12\Rightarrow3x=n\pi+(-1)^{n+1}\frac{\pi}{6}\Rightarrow x=\frac{n\pi}{3}+(-1)^{n+1}\frac{\pi}{18} \).

Teacher's Note:
1. When an equation contains a square root, squaring can add false roots.
2. Always check that the other side is non-negative.

Answer: (c) \( \frac13 \) or \( \frac53 \)

Explanation:
1. \( \cos^2\theta=\frac{1}{1+\tan^2\theta}=\frac{1}{1+\frac54}=\frac49 \), so \( \cos\theta=\pm\frac23 \).
2. If \( \cos\theta=\frac23 \), \( \text{versin}\,\theta=1-\frac23=\frac13 \).
3. If \( \cos\theta=-\frac23 \), \( \text{versin}\,\theta=1+\frac23=\frac53 \). The double result arises because tan²θ fixes only cos²θ.

Teacher's Note:
1. tan²θ determines only the magnitude of cos θ, so both signs are possible.
2. This explains the two values.

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

Explanation:
1. \( \sin3\theta-\sin2\theta=2\cos\frac{5\theta}{2}\sin\frac{\theta}{2}=0 \).
2. \( \sin\frac{\theta}{2}=0\Rightarrow\theta=2n\pi \).
3. \( \cos\frac{5\theta}{2}=0\Rightarrow\frac{5\theta}{2}=(2n+1)\frac{\pi}{2}\Rightarrow\theta=\frac{(2n+1)\pi}{5} \).

Teacher's Note:
1. The odd multiples of π/5 include the odd multiples of π but never 2nπ, so both families are needed.