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

The number of solutions of the equation $\sec x = 1 + \cos x + \cos^2 x + \dots \infty$ in the interval $x \in [-50 \pi, 50 \pi]$ is -

Solve $\cos x = \frac{1}{2}$

The most general value of $\theta$ which satisfies both the equations $\tan \theta = -1$ and $\cos \theta = \frac{1}{\sqrt{2}}$ is

The general solution of $\tan 3x = 1$ is

If the solutions for $\theta$ of $\cos p\theta + \cos q\theta = 0$ with $p > 0, q > 0$ are in an $A.P.$,then the numerically smallest common difference of the $A.P.$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module