If 3 sin x + cos x - 2 = (y - 1) ^ 2 for 0 ≤ x ≤ 8π , then the number of values of the pair (x, y) is equal to

Correct answer: 4 max . ( 3 sin x + cos x) = 3 ^ 2 + 1 ^ 2 = 2 and (y - 1) ^ 2 ≥ 0 So, LHS ≤ 0 and RHS ≥ 0 So, the possibility of equality is only when LHS = RHS = 0 RHS = 0 ⇒ y = 1 LHS = 0 ⇒ 3 sin x + cos x = 2 ⇒ 3 2 sin x + (1)/(2)cos x = 1 ⇒ (x - (π)/(3)) = 1 ⇒ x - (π)/(3) = 2nπ ⇒ x = 2nπ + (π)/(3) 0 ≤ 2nπ + (π)/(3) ≤ 8π ⇒ n = 0, 1, 2, 3 ⇒ There are four values of x and hence four pairs of (x, y)

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Trigonometric Equations

NTA Abhyas 2020 · Mathematics · Trigonometric Equations

If 3 sin x + cos x - 2 = (y - 1) ^ 2 for 0 ≤ x ≤ 8π , then the...

\sqrt{3}sin x + cos x - 2 = {\left(y - 1\right)}^{2}

for

0 \le x \le 8\pi , then the number of values of the pair \left(x, y\right)

is equal to

Correct answer

Step-by-step solution

max . \left(\sqrt{3}sin x + cos x\right) = \sqrt{{\sqrt{3}}^{2} + {1}^{2}} = 2

and

{\left(y - 1\right)}^{2} \ge 0

So, LHS

\le 0

and RHS

\ge 0

So, the possibility of equality is only when
LHS = RHS

= 0

RHS

= 0 ⇒ y = 1

LHS

= 0 ⇒ \sqrt{3}sin x + cos x = 2

⇒ \dfrac{\sqrt{3}}{2}sin x + \dfrac{1}{2}cos x = 1

⇒ \cos \left(x - \dfrac{\pi}{3}\right) = 1

⇒ x - \dfrac{\pi}{3} = 2n\pi ⇒ x = 2n\pi + \dfrac{\pi}{3}

0 \le 2n\pi + \dfrac{\pi}{3} \le 8\pi ⇒ n = 0, 1, 2, 3

⇒

There are four values of

x

and hence four pairs of

\left(x, y\right)

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