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 $[0, 2 π]$ is [2007]

Question

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 $[0, 2 π]$ is [2007]

Smart Achievers · Accepted Answer

(3)

$2 \sin^{2} θ - \cos 2 θ = 0$

$\Rightarrow \cos 2 θ = \frac{1}{2} \Rightarrow 2 θ = \frac{π}{3}, \frac{5 π}{3}, \frac{7 π}{3}, \frac{11 π}{3}$

$\Rightarrow θ = \frac{π}{6}, \frac{5 π}{6}, \frac{7 π}{6}, \frac{11 π}{6} . . . (i)$

$where θ \in [0, 2 π]$

$Also, 2 \cos^{2} θ - 3 \sin θ = 0$

$\Rightarrow 2 \sin^{2} θ + 3 \sin θ - 2 = 0$

$\Rightarrow (2 \sin θ - 1) (\sin θ + 2) = 0 \Rightarrow \sin θ = \frac{1}{2}$ $[∵ \sin θ \neq - 2]$

$\Rightarrow θ = \frac{π}{6}, \frac{5 π}{6} . . . (i i)$

$where θ \in [0, 2 π]$

$Combining (i) and (ii), we get θ = \frac{π}{6}, \frac{5 π}{6}$

$Hence, there are two solutions.$

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