If $\sin \theta + \sqrt{3} \cos \theta = 6x - x^2 - 11$,where $x \in R$ and $0 \le \theta \le 2\pi$,then the equation has a solution for:
- Aone value of $x$
- Btwo values of $x$
- Cinfinite values of $x$
- Dno value of $x$
Explore More
Similar Questions
If $A+B+C=\pi$ and $\cos B=\cos A \cos C$,then $\tan A \tan C=$
If $\sin^2 \theta = \frac{x^2 + y^2 + 1}{2x}$,then $x$ must be
Easy
View SolutionThe maximum value of $12 \sin \theta - 9 \sin^2 \theta$ is
If $A, B, C$ are the angles of a triangle,then the maximum value of $(\sin A + \sin B - \cos C)$ is-
The minimum value of $2^{\sin x} + 2^{\cos x}$ is
DifficultJEE MAIN 2020
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo