JEE Mathematics Trigonometric Equations MCQs Set 02

Answer: (a) \( x=\frac\pi4,\ y=1 \)

Explanation:
1. \( (\sin x+\cos x)^2\le2 \), so \( y+\frac1y\le2 \).
2. The expression \( y+\frac1y \) is ≥ 2 for y > 0 and ≤ −2 for y < 0; it must be non-negative, so y > 0 and \( y+\frac1y=2 \), giving y = 1.
3. Then \( \sin x+\cos x=\sqrt2 \), i.e. \( x=\frac\pi4 \).

Teacher's Note:
1. The maximum of sin x + cos x is √2, which bounds y + 1/y from above.
2. AM-GM gives y + 1/y ≥ 2 for positive y.

Answer: (b) \( \frac\pi4 \)

Explanation:
1. The equation is \( 2\sqrt2\,\sin x\,|\cos x|=1 \), so sin x > 0 and x ∈ (0, π).
2. If cos x > 0 (x < π/2): \( \sqrt2\sin2x=1 \), so \( 2x=\frac\pi4,\frac{3\pi}4 \), x = π/8, 3π/8.
3. If cos x < 0: \( -\sqrt2\sin2x=1 \), so \( 2x=\frac{5\pi}4,\frac{7\pi}4 \), x = 5π/8, 7π/8.
4. The roots π/8, 3π/8, 5π/8, 7π/8 are in AP with difference π/4.

Teacher's Note:
1. Replace √(8cos²x) by 2√2|cos x| and treat the sign of cos x separately.
2. sin x must be positive since the left side equals 1.

Answer: (a) \( \left(0,\frac\pi6\right)\cup\left(\frac{5\pi}6,2\pi\right) \)

Explanation:
1. \( (2\sin\theta-1)(\sin\theta-2)>0 \). Since \( \sin\theta-2<0 \) always, we need \( 2\sin\theta-1<0 \), i.e. \( \sin\theta<\frac12 \).
2. On (0, 2π), sin θ < 1/2 for \( \theta\in\left(0,\frac\pi6\right)\cup\left(\frac{5\pi}6,2\pi\right) \).

Teacher's Note:
1. Factorise the quadratic in sin θ.
2. The factor (sin θ − 2) is always negative.

Answer: (c) 2

Explanation:
1. Multiply by cos x (cos x ≠ 0): \( \sin x+1=2\cos^2x=2-2\sin^2x \), so \( 2\sin^2x+\sin x-1=0 \) and \( (2\sin x-1)(\sin x+1)=0 \).
2. sin x = −1 gives x = 3π/2, where cos x = 0 and the equation is undefined: rejected.
3. sin x = 1/2 gives x = π/6 and 5π/6: two solutions.

Teacher's Note:
1. Discard roots where the original functions are undefined.
2. Remember tan x and sec x are undefined where cos x = 0.

Answer: (b) \( \left\{\frac\pi3,\frac{5\pi}3\right\} \)

Explanation:
1. 2cos x − 1 = 0 gives \( \cos x=\frac12 \), so x = π/3 or 5π/3 in [0, 2π].
2. 3 + 2cos x = 0 gives \( \cos x=-\frac32 \), impossible since |cos x| ≤ 1.

Teacher's Note:
1. Set each factor to zero and check feasibility.
2. cos x cannot be less than −1.

Answer: (a) 0

Explanation:
1. Right side: \( \frac{\sin x-\cos x}{\sin x\cos x} \). Dividing by (sin x − cos x) ≠ 0 and putting \( p=\sin x\cos x \): \( 1+p-p^2=\frac1p \).
2. This gives \( p^3-p^2-p+1=(p-1)^2(p+1)=0 \), so p = 1 or p = −1.
3. But |p| = |sin 2x|/2 ≤ 1/2, so there is no solution.

Teacher's Note:
1. Factor out (sin x − cos x) and express the rest in terms of sin x cos x.
2. The bound |sin x cos x| ≤ 1/2 decides the matter.

Answer: (c) 3

Explanation:
1. Let \( f(x)=x+2\tan x-\frac\pi2 \); f′ = 1 + 2sec²x > 0, so f is increasing on each interval where tan x is defined.
2. On [0, π/2): f goes from −π/2 to +∞: 1 root. On (π/2, 3π/2): f goes from −∞ to +∞: 1 root.
3. On (3π/2, 2π]: f goes from −∞ to \( 2\pi-\frac\pi2>0 \): 1 root. Total 3.

Teacher's Note:
1. f is increasing on each branch of tan x, so count sign changes branch by branch.
2. Check the end values on every branch.

Answer: (a) \( n\pi+\frac{7\pi}6,\ n\in I \)

Explanation:
1. \( \tan\theta=\frac1{\sqrt3}\Rightarrow\theta=n\pi+\frac\pi6 \). For these values \( \sin2\theta=\sin\frac\pi3=\frac{\sqrt3}2 \), so both conditions hold.
2. Since \( n\pi+\frac\pi6=(n-1)\pi+\frac{7\pi}6 \), the set equals \( m\pi+\frac{7\pi}6,\ m\in I \).

Teacher's Note:
1. The condition tan θ = 1/√3 already fixes θ modulo π.
2. Re-index the integer to compare with 7π/6.

Answer: (a) nil

Explanation:
1. The left side lies in [0, 1], so \( 2[x]\in[0,1] \) and since 2[x] is an even integer, 2[x] = 0, i.e. [x] = 0 and x ∈ [0, 1).
2. Then |cos x| = 0, but cos x = 0 only at x = π/2 ≈ 1.57, which is outside [0, 1).
3. So there is no solution.

Teacher's Note:
1. 2[x] is always an even integer.
2. Check whether cos x can vanish for x in [0, 1).

Answer: (a) 0

Explanation:
1. \( \sin x\sin^2\frac x2=2\sin^3\frac x2\cos\frac x2 \). Let \( u=\frac x2\in\left[-\frac\pi2,\frac\pi2\right] \) and \( g(u)=2\sin^3u\cos u \).
2. \( g'(u)=2\sin^2u(3\cos^2u-\sin^2u)=0 \) gives \( \tan^2u=3 \), u = ±π/3.
3. The minimum is \( g\left(-\frac\pi3\right)=-2\cdot\frac{3\sqrt3}8\cdot\frac12=-\frac{3\sqrt3}8\approx-0.65>-1 \), so 1 + g(u) > 0: no solutions.

Teacher's Note:
1. Rewrite as a function of x/2 and find its minimum.
2. The product can never reach −1.

Answer: (d) none of these

Explanation:
1. \( \sin\alpha\cos\alpha(\cos^2\alpha-\sin^2\alpha)>0\Rightarrow\frac12\sin2\alpha\cos2\alpha>0\Rightarrow\sin4\alpha>0 \).
2. For α in (0, π), 4α ∈ (0, 4π) and sin 4α > 0 on (0, π) and (2π, 3π).
3. So \( \alpha\in\left(0,\frac\pi4\right)\cup\left(\frac\pi2,\frac{3\pi}4\right) \), which is not any of (a)-(c).

Teacher's Note:
1. Factor out sin α cos α to reach sin 4α > 0.
2. Remember the range of α is (0, π), so 4α covers two positive intervals.

Answer: (d) 2 or 6

Explanation:
1. Put \( \sin\theta=1-2\cos\theta \) into \( \sin^2\theta+\cos^2\theta=1 \): \( 5\cos^2\theta-4\cos\theta=0 \), so cos θ = 0 or 4/5.
2. cos θ = 0: sin θ = 1, value \( 0+6=6 \). cos θ = 4/5: sin θ = −3/5, value \( \frac{28}5-\frac{18}5=2 \).

Teacher's Note:
1. Eliminate sin θ using the given equation.
2. Evaluate the expression for each root.

Answer: (b) \( -\frac\pi2,\ 0,\ \frac\pi2 \)

Explanation:
1. Since |cos x| ≤ 1, \( \cos^7x\le\cos^2x \), and \( \sin^4x\le\sin^2x \). So \( \cos^7x+\sin^4x\le1 \).
2. Equality needs cos x = 0 or 1 and sin x ∈ {0, ±1}.
3. Within (−π, π): x = 0 (cos x = 1) and x = ±π/2 (cos x = 0).

Teacher's Note:
1. Compare each term with the square of the same function.
2. Find where both inequalities become equalities.

Answer: (d) A is false, but R is true.

Explanation:
1. Each sin²θᵢ = 0, so every cos θᵢ = ±1: R is true because the product of numbers ±1 is ±1.
2. For the sum to be n − 4, exactly two of the cosines must be −1 (n − 2k = n − 4, so k = 2). The number of such choices is \( \binom n2=\frac{n(n-1)}2 \), not n(n − 1): A is false.

Teacher's Note:
1. Every cosine is ±1.
2. Count how many ways two cosines out of n can be −1.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. \( \cos(\sin x)=\sin\left(\frac\pi2-\sin x\right) \), so the equation requires \( \cos x=\frac\pi2-\sin x+2k\pi \) or \( \cos x=\frac\pi2+\sin x+2k\pi \) (i.e. cos x + sin x or cos x − sin x equals π/2 + 2kπ).
2. The left sides lie in [−√2, √2] by R, while \( \frac\pi2\approx1.571>\sqrt2\approx1.414 \) and the other values π/2 + 2kπ are even further out. No solution: A is true, and R is the reason.

Teacher's Note:
1. Convert the cosine into a sine of the complementary angle.
2. Compare π/2 with √2.