If x is real and x+frac{1}{x}=2 cos theta, th | vedclass

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

Question

(C) Given the equation $x+\frac{1}{x}=2 \cos 	heta$.Multiplying by $x$,we get $x^2 - 2x \cos 	heta + 1 = 0$.Since $x$ is a real number,the discrimin

Vedclass · Accepted Answer

$\pm 1$

Vedclass · Answer

(C) Given the equation $x+\frac{1}{x}=2 \cos 	heta$.Multiplying by $x$,we get $x^2 - 2x \cos 	heta + 1 = 0$.Since $x$ is a real number,the discriminant $D$ of this quadratic equation must be greater than or equal to $0$.$D = b^2 - 4ac = (-2 \cos 	heta)^2 - 4(1)(1) = 4 \cos^2 	heta - 4$.Setting $D \geq 0$,we have $4 \cos^2 	heta - 4 \geq 0$,which implies $\cos^2 	heta \geq 1$.Since the range of $\cos 	heta$ is $[-1, 1]$,the only possible values satisfying $\cos^2 	heta \geq 1$ are $\cos 	heta = 1$ or $\cos 	heta = -1$.Thus,$\cos 	heta = \pm 1$.

If $x$ is real and $x+\frac{1}{x}=2 \cos \theta,$ then $\cos \theta=$

Similar Questions

$\log _{\frac{1}{8}\csc^2 \frac{\pi}{8}} \sin^2 \frac{3\pi}{8}$ equals to

If $\tan^2 \theta = 2\tan^2 \phi + 1$,then $\cos 2\theta + \sin^2 \phi$ equals

$\frac{\cos 17^\circ + \sin 17^\circ}{\cos 17^\circ - \sin 17^\circ} = $

$\tan 1^\circ \tan 2^\circ \tan 3^\circ \tan 4^\circ \dots \tan 89^\circ = $

The value of $\cos 24^{\circ} + \cos 5^{\circ} + \cos 300^{\circ} + \cos 175^{\circ} + \cos 204^{\circ}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module