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

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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

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