JEE Mathematics Trigonometric Equations MCQs Set 11

Answer: (a) \( 2\pi \)

Explanation:
1. The equation becomes \( 2\sin^2\theta-3\sin\theta-2=0 \), i.e. \( (\sin\theta-2)(2\sin\theta+1)=0 \), so \( \sin\theta=-\dfrac12 \).
2. In \( [-2\pi,2\pi] \): \( \theta=-\dfrac\pi6,-\dfrac{5\pi}6,\dfrac{7\pi}6,\dfrac{11\pi}6 \), with sum \( 2\pi \). Answer: (a)

Teacher's Note:
1. Reduce to a quadratic in \( \sin\theta \).

Answer: (b) 5

Explanation:
1. \( 1+\sin^4x\in[1,2] \) and \( \cos^23x\in[0,1] \), so equality needs \( \sin^4x=0 \) and \( \cos^23x=1 \).
2. \( x=k\pi \) gives \( \cos^23x=1 \); in the interval \( k=-2,-1,0,1,2 \), i.e. 5 solutions. Answer: (b)

Teacher's Note:
1. Compare the ranges of the two sides.

Answer: (d) \( [2,6] \)

Explanation:
1. \( 1-2\sin^2x+\alpha\sin x=2\alpha-7 \) gives \( 2(\sin^2x-4)-\alpha(\sin x-2)=0 \), i.e. \( (\sin x-2)(2\sin x+4-\alpha)=0 \).
2. As \( \sin x\ne2 \), \( \sin x=\dfrac{\alpha-4}2\in[-1,1] \), hence \( \alpha\in[2,6] \). Answer: (d)

Teacher's Note:
1. Factor the quadratic in \( \sin x \).

Answer: (a) have infinitely many solutions if \( \theta\in\left(\dfrac\pi2,\dfrac{2\pi}3\right) \) and has a unique solution if \( \theta\in\left(\pi,\dfrac{7\pi}6\right) \)

Explanation:
1. For \( \theta\in\left(\dfrac\pi2,\dfrac{2\pi}3\right) \): \( [\sin\theta]=0 \), \( [-\cos\theta]=0 \), \( [\cot\theta]=-1 \). The determinant \( [\sin\theta]-[-\cos\theta][\cot\theta]=0 \) and both rows are multiples of \( (0,0) \) or the same equation, so infinitely many solutions.
2. For \( \theta\in\left(\pi,\dfrac{7\pi}6\right) \): \( [\sin\theta]=-1 \), \( [-\cos\theta]=0 \), \( [\cot\theta]=n\ge1 \). The determinant is \( -1\ne0 \), so the solution is unique. Answer: (a)

Teacher's Note:
1. Evaluate the greatest-integer entries and the determinant in each interval.

Answer: (a) 2

Explanation:
1. \( \sin x+\sin3x=2\sin2x\cos x \), so the equation is \( \sin2x(2\cos x-1)=0 \).
2. \( \sin2x=0 \) gives \( x=0 \) (as \( x\lt\pi/2 \)); \( \cos x=\dfrac12 \) gives \( x=\dfrac\pi3 \). Two values. Answer: (a)

Teacher's Note:
1. Combine \( \sin x+\sin3x \).

Answer: (c) \( \dfrac\pi2 \)

Explanation:
1. Let \( c=\cos^22\theta \): \( 1-c+c^2=\dfrac34 \) gives \( \left(c-\dfrac12\right)^2=0 \), so \( \cos2\theta=\pm\dfrac1{\sqrt2} \).
2. For \( 2\theta\in(0,\pi) \): \( 2\theta=\dfrac\pi4,\dfrac{3\pi}4 \), so \( \theta=\dfrac\pi8,\dfrac{3\pi}8 \) with sum \( \dfrac\pi2 \). Answer: (c)

Teacher's Note:
1. Write everything in \( \cos^22\theta \).

Answer: (b) \( \dfrac{13}9 \)

Explanation:
1. \( \cos(\tfrac\pi6+x)\cos(\tfrac\pi6-x)=\cos^2\tfrac\pi6-\sin^2x=\dfrac34-\sin^2x \), so the bracket is \( \dfrac14-\sin^2x=\cos^2x-\dfrac34 \).
2. Then \( 8\cos x\left(\cos^2x-\dfrac34\right)=2(4\cos^3x-3\cos x)=2\cos3x=1 \), so \( \cos3x=\dfrac12 \), \( 3x=\dfrac\pi3,\dfrac{5\pi}3,\dfrac{7\pi}3 \), \( x=\dfrac\pi9,\dfrac{5\pi}9,\dfrac{7\pi}9 \). Sum \( =\dfrac{13\pi}9 \). Answer: (b)

Teacher's Note:
1. Reduce to \( \cos3x \).

Answer: (d) \( -\dfrac79 \)

Explanation:
1. Put \( y=\cos2x \): \( 5\left(\dfrac{1-y}{1+y}-\dfrac{1+y}2\right)=2y+9 \), which simplifies to \( 9y^2+42y+13=0 \), i.e. \( (3y+1)(3y+13)=0 \).
2. Since \( |y|\le1 \), \( \cos2x=-\dfrac13 \), so \( \cos4x=2\cos^22x-1=\dfrac29-1=-\dfrac79 \). Answer: (d)

Teacher's Note:
1. Express in \( \cos2x \).

Answer: (c) 7

Explanation:
1. Grouping: \( 2\cos2x\cos x+2\cos3x\cos x=0 \), so \( 2\cos x(\cos2x+\cos3x)=0 \), i.e. \( 4\cos x\cos\dfrac{5x}2\cos\dfrac x2=0 \).
2. \( \cos x=0 \): \( \dfrac\pi2,\dfrac{3\pi}2 \). \( \cos\dfrac{5x}2=0 \): \( \dfrac\pi5,\dfrac{3\pi}5,\pi,\dfrac{7\pi}5,\dfrac{9\pi}5 \). \( \cos\dfrac x2=0 \): \( x=\pi \) (repeated). Total distinct values: 7. Answer: (c)

Teacher's Note:
1. Pair \( x\leftrightarrow3x \) and \( 2x\leftrightarrow4x \)... then factor.

Answer: (a) \( \dfrac{2\pi}9,\dfrac\pi4,\dfrac{4\pi}9,\dfrac\pi2,\dfrac{3\pi}4,\dfrac{8\pi}9 \)

Explanation:
1. \( \sin\theta+\sin7\theta=2\sin4\theta\cos3\theta \), so the equation is \( \sin4\theta(1+2\cos3\theta)=0 \).
2. \( \sin4\theta=0 \): \( \theta=\dfrac\pi4,\dfrac\pi2,\dfrac{3\pi}4 \). \( \cos3\theta=-\dfrac12 \) with \( 3\theta\in(0,3\pi) \): \( 3\theta=\dfrac{2\pi}3,\dfrac{4\pi}3,\dfrac{8\pi}3 \), \( \theta=\dfrac{2\pi}9,\dfrac{4\pi}9,\dfrac{8\pi}9 \). Answer: (a)

Teacher's Note:
1. Combine the outer terms.

Answer: (d) 4

Explanation:
1. \( (2\sin x-1)(\sin x+3)=0 \), so \( \sin x=\dfrac12 \).
2. In \( [0,3\pi] \): \( \dfrac\pi6,\dfrac{5\pi}6,\dfrac{13\pi}6,\dfrac{17\pi}6 \), i.e. 4 values. Answer: (d)

Teacher's Note:
1. Factor the quadratic in \( \sin x \).

Answer: (b) \( -\dfrac{4+\sqrt7}3 \)

Explanation:
1. With \( t=\tan\dfrac x2 \): \( \dfrac{1-t^2+2t}{1+t^2}=\dfrac12 \), so \( 3t^2-4t-1=0 \), \( t=\dfrac{2\pm\sqrt7}3 \). As \( 0\lt\dfrac x2\lt\dfrac\pi2 \), \( t=\dfrac{2+\sqrt7}3 \).
2. \( \tan x=\dfrac{2t}{1-t^2}=-\dfrac{4+\sqrt7}3 \) (numerically \( -2.215 \), consistent with \( x\approx114^\circ \)). Answer: (b)

Teacher's Note:
1. Use the half-angle substitution.

Answer: (c) no solution

Explanation:
1. \( -\sqrt{a^2+b^2}\le a\sin x+b\cos x\le\sqrt{a^2+b^2} \) for all \( x \).
2. Since \( |c|\gt\sqrt{a^2+b^2} \), \( c \) lies outside this range, so there is no solution. Answer: (c)

Teacher's Note:
1. Compare \( c \) with the range of the left side.

Answer: (b) \( -\dfrac{24}{25} \)

Explanation:
1. \( (5\cos\alpha-3)(5\cos\alpha+4)=0 \); in the second quadrant \( \cos\alpha=-\dfrac45 \), \( \sin\alpha=\dfrac35 \).
2. \( \sin2\alpha=2\cdot\dfrac35\cdot\left(-\dfrac45\right)=-\dfrac{24}{25} \). Answer: (b)

Teacher's Note:
1. Factor the quadratic in \( \cos\theta \).