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JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Trigonometric Equations

The sum of all values of

θ \in [0, 2 π]

satisfying

2 sin^{2} θ = cos 2 θ

and

2 cos^{2} θ = 3 sin θ

is

(A) $\frac{π}{2}$
(B) $4 π$
(C) $\frac{5 π}{6}$
(D) $π$ ✓ Correct

Answer: (D)

Solution

From

2 sin^{2} θ = cos 2 θ = 1 - 2 sin^{2} θ

:

4 sin^{2} θ = 1 \Rightarrow sin θ = \pm \frac{1}{2} .

From

2 cos^{2} θ = 3 sin θ

:

2 (1 - sin^{2} θ) = 3 sin θ \Rightarrow 2 sin^{2} θ + 3 sin θ - 2 = 0 \Rightarrow (2 sin θ - 1) (sin θ + 2) = 0.

Since

sin θ = - 2

is impossible,

sin θ = \frac{1}{2}

, which also satisfies the first equation.

In

[0, 2 π]

,

sin θ = \frac{1}{2}

gives

θ = \frac{π}{6}, \frac{5 π}{6}

.

Sum

= \frac{π}{6} + \frac{5 π}{6} = π .

JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Single correct option

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