Statement -1: The number of common solutions | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) For Statement $-1$: Solve $2\sin^2	heta - \cos 2	heta = 0$.$2\sin^2	heta - (1 - 2\sin^2	heta) = 0$ $\Rightarrow 4\sin^2	heta = 1$ $\Rightarro

Vedclass · Accepted Answer

Statement $-1$ is true; Statement $-2$ is true; Statement $-2$ is not a correct explanation for Statement $-1$.

Vedclass · Answer

(B) For Statement $-1$: Solve $2\sin^2	heta - \cos 2	heta = 0$.$2\sin^2	heta - (1 - 2\sin^2	heta) = 0$ $\Rightarrow 4\sin^2	heta = 1$ $\Rightarrow \sin	heta = \pm \frac{1}{2}$.In $[0, 2\pi]$,$	heta \in \{\frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}\}$.Solve $2\cos^2	heta - 3\sin	heta = 0$.$2(1 - \sin^2	heta) - 3\sin	heta = 0 \Rightarrow 2\sin^2	heta + 3\sin	heta - 2 = 0$.$(2\sin	heta - 1)(\sin	heta + 2) = 0$.Since $\sin	heta = -2$ is impossible,$\sin	heta = \frac{1}{2}$,so $	heta \in \{\frac{\pi}{6}, \frac{5\pi}{6}\}$.Common solutions are $\{\frac{\pi}{6}, \frac{5\pi}{6}\}$,so Statement $-1$ is true.For Statement $-2$: Solve $2\cos^2	heta - 3\sin	heta = 0$ in $[0, \pi]$.As shown above,$\sin	heta = \frac{1}{2}$ gives $	heta = \frac{\pi}{6}, \frac{5\pi}{6}$. Both are in $[0, \pi]$.Thus,Statement $-2$ is true. However,Statement $-2$ describes the solution set of one equation,which does not explain why the common solutions of both equations are two. Thus,Statement $-2$ is not the correct explanation.

Similar Questions

Let $\frac{\sin A}{\sin B} = \frac{\sin (A-C)}{\sin (C-B)}$,where $A, B, C$ are angles of a triangle $ABC$. If the lengths of the sides opposite these angles are $a, b, c$ respectively,then:

In a $\Delta ABC$,$a = a_1 = 2$,$b = a_2$,$c = a_3$ such that $a_{p+1} = \frac{5^p}{3^{2-p}} a_p \left( 2^{2-p} - \frac{4p-2}{5^p} a_p \right)$ where $p = 1, 2$,then:

The number of solutions to $\sin(\pi \sin^2 \theta) + \sin(\pi \cos^2 \theta) = 2 \cos(\frac{\pi}{2} \cos \theta)$ satisfying $0 \leq \theta \leq 2\pi$ is

Statement-$I$: In the interval $[0, 2\pi]$,the number of common solutions of the equations $2 \sin^2 \theta - \cos 2\theta = 0$ and $2 \cos^2 \theta - 3 \sin \theta = 0$ is two.
Statement-$II$: The number of solutions of $2 \cos^2 \theta - 3 \sin \theta = 0$ in $[0, \pi]$ is two.

In $\triangle ABC$,if $3 \sin A + 4 \cos B = 6$ and $4 \sin B + 3 \cos A = 1$,then the $\angle C$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module