JEE Advanced 2007 Paper 1Question 1
The number of solutions of the pair of equations $$2\sin^2\theta - \cos 2\theta = 0$$ $$2\cos^2\theta - 3\sin\theta = 0$$ in the interval $[0, 2\pi]$ is
Trigonometric Ratio & Identities
Solution
Concept. Write both equations in $\sin\theta$ and take the common values. Formulas used. (1) $\cos2\theta=1-2\sin^2\theta$ and $\cos^2\theta=1-\sin^2\theta$Step 1 - First equation. By (1), $4\sin^2\theta-1=0$, so $\sin\theta=\pm\frac12$. Step 2 - Second equation. By (1), $2\sin^2\theta+3\sin\theta-2=0$, i.e. $(2\sin\theta-1)(\sin\theta+2)=0$. Only $\sin\theta=\frac12$ is possible. Step 3 - Common solutions. $\sin\theta=\frac12$ satisfies both: $\theta=\frac\pi6,\frac{5\pi}6$ in $[0,2\pi]$. Two solutions. Answer: (C)