The number of real values of x in [0, 2pi] - | vedclass

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Question

(C) Given the equation: $|\cos x|^{2\sin^2 x - 3\sin x + 1} = 1$. This equation holds if: $1)$ $|\cos x| = 1$ $2)$ The exponent is $0$ (provided the b

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) Given the equation: $|\cos x|^{2\sin^2 x - 3\sin x + 1} = 1$. This equation holds if: $1)$ $|\cos x| = 1$ $2)$ The exponent is $0$ (provided the base is not $0$). Case $1$: $|\cos x| = 1 \implies \cos x = 1$ or $\cos x = -1$. For $x \in [0, 2\pi]$,this gives $x = 0, \pi, 2\pi$. Case $2$: $2\sin^2 x - 3\sin x + 1 = 0$. $(2\sin x - 1)(\sin x - 1) = 0$. $\sin x = \frac{1}{2}$ or $\sin x = 1$. If $\sin x = \frac{1}{2}$,then $x = \frac{\pi}{6}, \frac{5\pi}{6}$. If $\sin x = 1$,then $x = \frac{\pi}{2}$. However,the domain is $x \in [0, 2\pi] - \{\frac{\pi}{2}, \frac{3\pi}{2}\}$. Thus,$x = \frac{\pi}{2}$ is excluded. Also,we must check if the base $|\cos x| = 0$ when the exponent is $0$. If $x = \frac{\pi}{2}$ or $\frac{3\pi}{2}$,$|\cos x| = 0$,but these are excluded from the domain. Combining the valid solutions: $x \in \{0, \pi, 2\pi, \frac{\pi}{6}, \frac{5\pi}{6}\}$. The total number of values is $5$.

The number of real values of $x \in [0, 2\pi] - \{\frac{\pi}{2}, \frac{3\pi}{2}\}$ satisfying the equation $|\cos x|^{2\sin^2 x - 3\sin x + 1} = 1$ is:

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