Consider the following lists: List-I List-II I x∈-2π3,2π3:cosx+sinx=1 P has two elements II x∈-5π18,5π18:3tan3x=1 Q has three elements III x∈-6π5,6π5:2cos2x=3 R has four elements IV x∈-7π4,7π4:sinx-cosx=1 S has five elements T has six elements The correct option is:

(B) I→P;II→P;III→T;IV→R

(A) I→P;II→S;III→P;IV→S

(B) I→P;II→P;III→T;IV→R

(C) I→Q;II→P;III→T;IV→S

(D) I→Q;II→S;III→P;IV→R

JEE Advanced · Mathematics · 8. Trigonometric Equations

Consider the following lists:

List- $I$ List- $II$

$(I)$ $\{x \in [- \frac{2 π}{3}, \frac{2 π}{3}] : \cos x + \sin x = 1\}$ $(P)$ has two elements

$(II)$ $\{x \in [- \frac{5 π}{18}, \frac{5 π}{18}] : \sqrt{3} \tan 3 x = 1\}$ $(Q)$ has three elements

$(III)$ $\{x \in [- \frac{6 π}{5}, \frac{6 π}{5}] : 2 \cos (2 x) = \sqrt{3}\}$ $(R)$ has four elements

$(IV)$ $\{x \in [- \frac{7 π}{4}, \frac{7 π}{4}] : \sin x - \cos x = 1\}$ $(S)$ has five elements

$(T)$ has six elements

The correct option is:

A $(I) \to (P); (II) \to (S); (III) \to (P); (IV) \to (S)$
B $(I) \to (P); (II) \to (P); (III) \to (T); (IV) \to (R)$
C $(I) \to (Q); (II) \to (P); (III) \to (T); (IV) \to (S)$
D $(I) \to (Q); (II) \to (S); (III) \to (P); (IV) \to (R)$

Verified Solution

Answer & Solution

Correct Answer

(B) $(I) \to (P); (II) \to (P); (III) \to (T); (IV) \to (R)$

Step-by-step Solution

Detailed explanation

Solving all questions one by one we get,
(i)

\{x \in [\frac{- 2 π}{3}, \frac{2 π}{3}], \cos x + \sin x = 1\}

\cos x + \sin x = 1

\Rightarrow \sin (\frac{π}{4} + x) = \frac{1}{\sqrt{2}}

\Rightarrow \frac{π}{4} + x = n π + {(- 1)}^{n} \frac{π}{4}

\Rightarrow x = n π + {(- 1)}^{n} \frac{π}{4} - \frac{π}{4}

So,

x \in \{0, \frac{π}{2}\}

∴ x

has

2

elements.

\to P

(ii)

\{x \in [\frac{- 5 π}{18}, \frac{5 π}{18}] : \sqrt{3} \tan 3 x = 1\}

Solving

\sqrt{3} \tan 3 x = 1

\Rightarrow \tan 3 x = \frac{1}{\sqrt{3}}

\Rightarrow 3 x = n π + \frac{π}{6}

\Rightarrow x = \frac{n π}{3} + \frac{π}{18}

So,

x \in \{\frac{π}{18}, \frac{- 5 π}{18}\}

∴ x

has

2

elements.

\to P

(iii)

\{x \in [\frac{- 6 π}{5}, \frac{6 π}{5}] : 2 \cos 2 x = \sqrt{3}\}

2 \cos 2 x = \sqrt{3}

\Rightarrow \cos 2 x = \frac{\sqrt{3}}{2}

\Rightarrow 2 x = 2 n π \pm \frac{π}{6}

\Rightarrow x = n π \pm \frac{π}{12}

So,

x \in \{\pm \frac{π}{12}, π \pm \frac{π}{12}, - π \pm \frac{π}{12}\}

∴ x

has

6

elements.

\to T

(iv)

\{x \in [\frac{- 7 π}{4}, \frac{7 π}{4}] : \sin x - \cos x = 1\}

\sin x - \cos x = 1

\Rightarrow \sin (x - \frac{π}{4}) = \frac{1}{\sqrt{2}}

\Rightarrow x - \frac{π}{4} = n π + {(- 1)}^{n} \frac{π}{4}

\Rightarrow x = n π + {(- 1)}^{n} \frac{π}{4} + \frac{π}{4}

So,

x \in \{\frac{π}{2}, \frac{- 3 π}{2}, π, - π\}

∴ x

has

4

elements.

\to R

∴

option

B

is correct.

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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Column I	Column II
A. Carbonate	P. Siderite
B. Sulphide	Q. Malachite
C. Hydroxide	R. Bauxite
D. Oxide	S. Calamine
	T. Argentite

	List- $I$		List- $II$
$(I)$	$\{x \in [- \frac{2 π}{3}, \frac{2 π}{3}] : \cos x + \sin x = 1\}$	$(P)$	has two elements
$(II)$	$\{x \in [- \frac{5 π}{18}, \frac{5 π}{18}] : \sqrt{3} \tan 3 x = 1\}$	$(Q)$	has three elements
$(III)$	$\{x \in [- \frac{6 π}{5}, \frac{6 π}{5}] : 2 \cos (2 x) = \sqrt{3}\}$	$(R)$	has four elements
$(IV)$	$\{x \in [- \frac{7 π}{4}, \frac{7 π}{4}] : \sin x - \cos x = 1\}$	$(S)$	has five elements
		$(T)$	has six elements