The real roots of the equation cos^7x + sin^4 | vedclass

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(B) Given equation: $\cos^7x + \sin^4x = 1$We know $\sin^4x = (1 - \cos^2x)^2 = 1 - 2\cos^2x + \cos^4x$.Substituting this into the equation: $\cos^7x

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

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(B) Given equation: $\cos^7x + \sin^4x = 1$We know $\sin^4x = (1 - \cos^2x)^2 = 1 - 2\cos^2x + \cos^4x$.Substituting this into the equation: $\cos^7x + 1 - 2\cos^2x + \cos^4x = 1$$\cos^7x + \cos^4x - 2\cos^2x = 0$$\cos^2x(\cos^5x + \cos^2x - 2) = 0$Case $1$: $\cos^2x = 0 \implies \cos x = 0 \implies x = \pm \frac{\pi}{2}$.Case $2$: $\cos^5x + \cos^2x - 2 = 0$.Let $f(t) = t^5 + t^2 - 2$ where $t = \cos x$. Since $t \in [-1, 1]$,the maximum value of $t^5 + t^2$ is $1^5 + 1^2 = 2$.Thus,$t^5 + t^2 - 2 = 0$ only when $t = 1$,which means $\cos x = 1$.$\cos x = 1 \implies x = 0$.Combining the results,the roots in $(-\pi, \pi)$ are $\{ -\frac{\pi}{2}, 0, \frac{\pi}{2} \}$.

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

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