Trigonometric Equations

Q: The number of roots of the equation x + 2tanx = π/2 in the interval [0, 2π] is

(A) 1

(B) 2

(C) 3

(D) infinite

Sol. We have x + 2 tanx = π/2 or tan x = π/4 – x/2.

Now the graphs of the curve y = tanx and y = π/4- x/2, in the interval [0, 2π] intersect at three points.
The abscissa of these three points are the roots of the equation.

Hence (C) is the correct answer.

Q: The number of all possible 5-tuples (a₁, a₂, a₃, a₄, a₅ ) such that
a₁ + a₂ sinx + a₃ cosx + a₄ sin2x + a₅ cos2x = 0 holds for all x is

(A) zero

(B) 1

(C) 2

(D) infinite

Sol. Since the equation a₁ + a₂ sinx + a₃ cosx + a₄ sin2x + a₅ cos2x = 0 holds for all values of x ,

a1+ a3 + a5= 0 ( on putting x = 0)

a1- a3+ a5= 0 ( on putting x = π)

⇒ a3 =0 and a1 + a5 = 0 . . . . (1)

Putting x = π/2 and 3π/2 , we get

a1+ a2- a5 = 0 and a1- a2- a5 = 0

⇒ a2 =0 and a1 –a5 =0 . . . . (2)

(1) and (2) give

a1 =a2 = a3 = a5 =0

The given equation reduces to a4sin2x = 0

This is true for all values of x , therefore a4 = 0

Hence a1 =a2 = a3 = a4 =a5 = 0.

Thus the number of 5-tuples is one.

Hence (B) is the correct answer.

Q: If [sin^-1cos^-1 sin^-1 tan^-1 x] = 1 , where [.] denotes the greatest integer function, then x belongs to the interval

(A) [ tan sin cos1 , tan sin cos sin1 ]

(B) [ tan sin cos1 , tan sin cos sin1 ]

(C) [1 , -1]

(D) [ sin cos tan1, sin cos sin tan1 ]

Q: The number of points inside or on the circle x² + y² = 4 satisfying
tan⁴x + cot⁴x + 1 = 3sin²y is

(A) one

(B) two

(C) four

(D) infinite

Sol. tan⁴x + cot⁴x + 1 = (tan²x – cot²x)² + 3 ≤ 3

3 sin²y  ≤ 3

⇒ tan² x = cot² x , sin²y = 1

⇒ tanx = ± 1, siny = ±1

⇒ x = ± π/4, ± 3π/4 , . . .

But x² ≤  4 ⇒ -2 ≤  x ≤  2 ⇒ x = ± π/4 only

Siny = ± 1 ⇒  y = ± π/2 ,  3π/2 , . . ..

But y² ≤ 4  ⇒ y = ± π/2  only.

So four solutions are possible.

Hence (C) is the correct answer.

Q: Indicate the relation which is false

(A) tan | tan^-1 x | = | x |

(B) cot | cot^-1 x | = x

(C) tan^-1 | tan x | = | x |

(D) sin | sin^-1 x | = | x |

Sol. Since $ \large | tan^{-1}x | = \left\{\begin{array}{ll} tan^{-1}x \; , if \; 0 \leq tan^{-1}x \le \pi/2 \\ -tan^{-1}x \; , if  \; -\pi/2 < tan^{-1}x < 0 \end{array} \right. $

$ \large = \left\{\begin{array}{ll} tan^{-1}x \; , if \; x \ge 0 \\ -tan^{-1}x \; , if x < 0 \end{array} \right. $

⇒ | tan^-1x | = tan^-1 |x| ∀ x ∈ R

⇒ tan| tan^-1x | = tan tan^-1 |x| = |x|

Hence (A) is the correct answer.

Likewise sin | sin^-1x| = sin sin^-1|x| = |x| ∀ |x| ≤ 1

Hence (D) is the correct answer.

| cot^-1x | = cot^-1x as 0 < | cot^-1x| < π ∀ x ∈ R

⇒ cot |cot^-1x| = cot cot^-1x = x .

Hence (B) is the correct answer.

Since | tanx | is not necessarily always equal to tan|x|

Hence tan^-1 | tanx | ≠ | x| .

Q: If k ≤ sin^-1x + cos^-1x + tan^-1x ≤ K , then

(A) k = 0, K = π

(B) k = 0, K = π/2

(C) k = π/2, K = π

(D) None of these