Statement-1: The number of common solutions of the trigonometric equations 2^2-2=0 and 2^2…

Question

Statement-1: The number of common solutions of the trigonometric equations $2\sin^2	heta-\cos2	heta=0$ and $2\cos^2	heta-3\sin	heta=0$ in the interval $[0,2\pi]$ is two.
Statement-2: The number of solutions of the equation $2\cos^2	heta-3\sin	heta=0$ in the interval $[0,\pi]$ is two.

Accepted Answer

$$2\sin^2\theta-\cos2\theta=0 \implies 4\sin^2\theta=1 \implies \sin\theta=\pm\frac12 \implies \theta=\frac\pi4,\frac{3\pi}4,\frac{5\pi}4,\frac{7\pi}4 \text{ or } \frac\pi6,\frac{5\pi}6,\frac{7\pi}6,\frac{11\pi}6$$

*(the relevant family from this equation is $\theta=\tfrac\pi6,\tfrac{5\pi}6,\tfrac{7\pi}6,\tfrac{11\pi}6$)*

$$2\cos^2\theta-3\sin\theta=0 \implies 2\sin^2\theta-3\sin\theta+... \implies (2\sin\theta-1)(\sin\theta+2)=0 \implies \sin\theta=\frac12 \implies \theta=\frac\pi6,\frac{5\pi}6$$

(within $[0,\pi]$ --- 2 solutions, confirming Statement-2). The common solutions of both original equations over $[0,2\pi]$ are $\theta=\tfrac\pi6,\tfrac{5\pi}6$ --- 2 solutions, confirming Statement-1. Both statements are true, but Statement-2 (about a different, single equation) is not a direct explanation of Statement-1 (about the intersection of two equations).

Solutions · 1