The real roots of the equation cos^7x, | vedclass

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

Question

$\cos ^{7} x+\sin ^{4} x=1$

$\cos ^{7} x=1-\sin ^{4} x=\left(1-\sin ^{2} x\right)\left(1+\sin ^{2} x\right)$

$\cos ^{7} x=\cos ^{2} x\left(1+\si

Vedclass · Accepted Answer

$\{ - \frac{\pi }{2}\,,\,0\,,\,\frac{\pi }{2} \}$

Vedclass · Answer

$\cos ^{7} x+\sin ^{4} x=1$

$\cos ^{7} x=1-\sin ^{4} x=\left(1-\sin ^{2} x\right)\left(1+\sin ^{2} x\right)$

$\cos ^{7} x=\cos ^{2} x\left(1+\sin ^{2} x\right)$

$\cos ^{7}(x)-\cos ^{2} x\left(2-\cos ^{2} x\right)=0$

$\cos ^{2} x\left(\cos ^{5} x+\cos ^{2}(x)-2\right)=0$

$\cos ^{2}(x)=0$ implies

$x=\frac{\pm \pi}{2}$

And

$\cos ^{5} x+\cos ^{2} x-2=0$ implies

$\cos (x)=1$

$x=0$

Hence

$x \in\left\{\frac{-\pi}{2}, 0, \frac{\pi}{2}\right\}$

The real roots of the equation $cos^7x\, +\, sin^4x\, =\, 1$ in the interval $(-\pi, \pi)$ are

Similar Questions

One of the solutions of the equation $8 \sin ^3 \theta-7 \sin \theta+\sqrt{3} \cos \theta=0$ lies in the interval

If $\sec x\cos 5x + 1 = 0$, where $0 < x < 2\pi $, then $x =$

If ${\sin ^2}\theta = \frac{1}{4},$ then the most general value of $\theta $ is

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

Let $f(x)=\cos 5 x+A \cos 4 x+B \cos 3 x$ $+C \cos 2 x+D \cos x+E$, and

$T=f(0)-f\left(\frac{\pi}{5}\right)+f\left(\frac{2 \pi}{5}\right)-f\left(\frac{3 \pi}{5}\right)+\ldots+f\left(\frac{8 \pi}{5}\right)-f\left(\frac{9 \pi}{5}\right) \text {. }$Then, $T$