The number of solutions of the pair of equations \[ \begin{aligned} & 2 \sin ^{2} \theta-\cos 2 \theta=0 \\ & 2 \cos ^{2} \theta-3 \sin \theta=0 \end{aligned} \] in the interval $[0,2 \pi]$ is

JEE Advanced · 2007 · 08 Apr 2007 · Paper 1 · Trigonometric Equations

The number of solutions of the pair of equations

2 sin^{2} θ - cos 2 θ = 0 2 cos^{2} θ - 3 sin θ = 0

in the interval

[0, 2 π]

Answer: (C)

From the first equation:

2 sin^{2} θ - cos 2 θ = 0

2 sin^{2} θ - (1 - 2 sin^{2} θ) = 0

4 sin^{2} θ = 1

sin^{2} θ = \frac{1}{4}

sin θ = \pm \frac{1}{2}

From the second equation:

2 cos^{2} θ = 3 sin θ

2 (1 - sin^{2} θ) = 3 sin θ

2 sin^{2} θ + 3 sin θ - 2 = 0

(2 sin θ - 1) (sin θ + 2) = 0

sin θ = \frac{1}{2}

(since

sin θ = - 2

is impossible)

Therefore,

sin θ = \frac{1}{2}

, which gives

θ = \frac{π}{6}, \frac{5 π}{6}

[0, 2 π]

.

There are two solutions.

JEE Advanced · 2007 · 08 Apr 2007 · Paper 1 · Single correct option