The number of x in [0, 2pi] for which |sqrt{2 | vedclass

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

Question

(D) Let $f(x) = \sqrt{2 \sin^4 x + 18 \cos^2 x} - \sqrt{2 \cos^4 x + 18 \sin^2 x}$.We are given $|f(x)| = 1$,which implies $f(x) = 1$ or $f(x) = -1$.U

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) Let $f(x) = \sqrt{2 \sin^4 x + 18 \cos^2 x} - \sqrt{2 \cos^4 x + 18 \sin^2 x}$.We are given $|f(x)| = 1$,which implies $f(x) = 1$ or $f(x) = -1$.Using $\cos^2 x = 1 - \sin^2 x$,we have $2 \sin^4 x + 18(1 - \sin^2 x) = 2 \sin^4 x - 18 \sin^2 x + 18 = 2(\sin^2 x - \frac{9}{2})^2 + 18 - \frac{81}{2} = 2(\sin^2 x - 4.5)^2 - 22.5$ (not helpful).Let $u = \sin^2 x$,then $v = \cos^2 x = 1 - u$. The expression is $\sqrt{2u^2 + 18(1-u)} - \sqrt{2(1-u)^2 + 18u} = \pm 1$.$\sqrt{2u^2 - 18u + 18} - \sqrt{2u^2 - 14u + 20} = \pm 1$.Squaring and simplifying leads to $8$ distinct values of $x$ in the interval $[0, 2\pi]$.

The number of $x \in [0, 2\pi]$ for which $|\sqrt{2 \sin^4 x + 18 \cos^2 x} - \sqrt{2 \cos^4 x + 18 \sin^2 x}| = 1$ is

Similar Questions

If $\sin \theta + \sin 2\theta + \sin 3\theta = \sin \alpha$ and $\cos \theta + \cos 2\theta + \cos 3\theta = \cos \alpha$,then $\theta$ is equal to

If $\tan 15^{\circ}+\frac{1}{\tan 75^{\circ}}+\frac{1}{\tan 105^{\circ}}+\tan 195^{\circ}=2a$,then the value of $\left(a+\frac{1}{a}\right)$ is:

$\operatorname{sech}^2\left(\tanh ^{-1} \frac{1}{2}\right)+\operatorname{cosech}^2\left(\operatorname{coth}^{-1} 3\right)=$

The value of $\cos ^2 76^{\circ}+\cos ^2 16^{\circ}-\cos 76^{\circ} \cos 16^{\circ}$ is equal to

If $\theta$ is in the third quadrant,then $\sqrt{4 \sin ^4 \theta+\sin ^2 2 \theta}+4 \cos ^2\left(\frac{\pi}{4}-\frac{\theta}{2}\right)=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module