The value of x, which satisfies the equation | vedclass

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

Question

(D) We know that $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,$\operatorname{cosec} 30^{\circ} = \frac{1}{\sin 30^{\circ}} = \frac{1}{1/2} = 2$,and $	an 30

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) We know that $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,$\operatorname{cosec} 30^{\circ} = \frac{1}{\sin 30^{\circ}} = \frac{1}{1/2} = 2$,and $	an 30^{\circ} = \frac{1}{\sqrt{3}}$.Substituting these values into the given equation:$2(2)^{2} + x(\frac{\sqrt{3}}{2})^{2} - \frac{3}{4}(\frac{1}{\sqrt{3}})^{2} = 10$$2(4) + x(\frac{3}{4}) - \frac{3}{4}(\frac{1}{3}) = 10$$8 + \frac{3x}{4} - \frac{1}{4} = 10$Multiply the entire equation by $4$ to eliminate the denominators:$32 + 3x - 1 = 40$$31 + 3x = 40$$3x = 40 - 31$$3x = 9$$x = 3$

The value of $x,$ which satisfies the equation $2 \operatorname{cosec}^{2} 30^{\circ} + x \sin^{2} 60^{\circ} - \frac{3}{4} \tan^{2} 30^{\circ} = 10$ is

Similar Questions

If $\tan \theta + \sec \theta = e^x$,then $\cos \theta$ equals

$\frac{{\cos 12^\circ - \sin 12^\circ }}{{\cos 12^\circ + \sin 12^\circ }} + \frac{{\sin 147^\circ }}{{\cos 147^\circ }} = $

For all $\theta,$ the value of $\sqrt{\frac{1+\sin \theta}{1-\sin \theta}}$ is:

If $\sin 3A = \cos (A - 26^{\circ})$,where $3A$ is an acute angle,then the value of $A$ is (in $^{\circ}$)

If $\frac{\cos \alpha}{\sin \beta} = n$ and $\frac{\cos \alpha}{\cos \beta} = m$,then the value of $\cos^2 \beta$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module