The general solution of cos 2x - 2 tan x + 2 | vedclass

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

Question

(D) Given equation: $\cos 2x - 2 	an x + 2 = 0$Using the identity $\cos 2x = \frac{1 - 	an^2 x}{1 + 	an^2 x}$,we get:$\frac{1 - 	an^2 x}{1 + 	an^

Vedclass · Accepted Answer

$n\pi + \frac{\pi}{4}, n \in Z$

Vedclass · Answer

(D) Given equation: $\cos 2x - 2 	an x + 2 = 0$Using the identity $\cos 2x = \frac{1 - 	an^2 x}{1 + 	an^2 x}$,we get:$\frac{1 - 	an^2 x}{1 + 	an^2 x} - 2 	an x + 2 = 0$Multiplying by $(1 + 	an^2 x)$:$(1 - 	an^2 x) - 2 	an x(1 + 	an^2 x) + 2(1 + 	an^2 x) = 0$$1 - 	an^2 x - 2 	an x - 2 	an^3 x + 2 + 2 	an^2 x = 0$$-2 	an^3 x + 	an^2 x - 2 	an x + 3 = 0$$2 	an^3 x - 	an^2 x + 2 	an x - 3 = 0$Let $	an x = t$,then $2t^3 - t^2 + 2t - 3 = 0$.By inspection,$t = 1$ is a root: $2(1)^3 - (1)^2 + 2(1) - 3 = 2 - 1 + 2 - 3 = 0$.Dividing by $(t - 1)$,we get $(t - 1)(2t^2 + t + 3) = 0$.The quadratic $2t^2 + t + 3$ has discriminant $D = 1^2 - 4(2)(3) = 1 - 24 = -23 Thus,$	an x = 1 = 	an \frac{\pi}{4}$.The general solution is $x = n\pi + \frac{\pi}{4}, n \in Z$.

The general solution of $\cos 2x - 2 \tan x + 2 = 0$ is

Similar Questions

If $\sec^2 \theta = \frac{4}{3}$,then the general value of $\theta$ is

If $\cos \theta + \cos 2\theta + \cos 3\theta = 0$,then the general value of $\theta$ is

If $6 \cos 2 \theta + 2 \cos^2 \left(\frac{\theta}{2}\right) + 2 \sin^2 \theta = 0$ for $-\pi < \theta < \pi$,then $\theta =$

The number of values of $x$ in the interval $[0, 3\pi]$ satisfying the equation $2 \sin^2 x + 5 \sin x - 3 = 0$ is

The solution of $3 \tan (A - 15^{\circ}) = \tan (A + 15^{\circ})$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module