The number of solutions of the equation 2 the | vedclass

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

Question

(A) The given equation is $2 	heta - \cos^{2} 	heta + \sqrt{2} = 0$.This can be rewritten as $\cos^{2} 	heta = 2 	heta + \sqrt{2}$.Let $f(	heta)

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) The given equation is $2 	heta - \cos^{2} 	heta + \sqrt{2} = 0$.This can be rewritten as $\cos^{2} 	heta = 2 	heta + \sqrt{2}$.Let $f(	heta) = \cos^{2} 	heta$ and $g(	heta) = 2 	heta + \sqrt{2}$.The function $f(	heta) = \cos^{2} 	heta$ is a periodic function with a range of $[0, 1]$.The function $g(	heta) = 2 	heta + \sqrt{2}$ is a straight line with a slope of $2$ and a $y$-intercept of $\sqrt{2} \approx 1.414$.Since the maximum value of $\cos^{2} 	heta$ is $1$,and for $	heta > 0$,$g(	heta) > \sqrt{2} > 1$,there are no solutions for $	heta > 0$.For $	heta < 0$,the line $g(	heta)$ intersects the curve $f(	heta)$ at exactly one point,as shown in the graph.

The number of solutions of the equation $2 \theta - \cos^{2} \theta + \sqrt{2} = 0$ is equal to

Similar Questions

If $\cot \frac{x}{2} - \operatorname{cosec} \frac{x}{2} = \cot x$,then the values of $x$ are

The solution of the equation $\sec \theta - \text{cosec} \theta = \frac{4}{3}$ is

The general solution of $\sin x - 3 \sin 2x + \sin 3x = \cos x - 3 \cos 2x + \cos 3x$ is

The solution set of the equation $\cos^2 2x + \sin^2 3x = 1$ is

The number of solutions in $[0, \frac{\pi}{2})$ of the equation $\cos 3x \cdot \tan 5x = \sin 7x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module