The general solution of sin^{2} x cdot sec x | vedclass

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(B) Given equation: $\sin^{2} x \sec x = 	an x - \sin x + 1$Multiply by $\cos x$ (where $\cos x 
eq 0$):$\sin^{2} x = \sin x - \sin x \cos x + \cos

Vedclass · Accepted Answer

$x = n \pi + (-1)^{n} \frac{\pi}{2}$ or $x = m \pi + \frac{3 \pi}{4}; m, n \in \mathbb{Z}$

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(B) Given equation: $\sin^{2} x \sec x = 	an x - \sin x + 1$Multiply by $\cos x$ (where $\cos x 
eq 0$):$\sin^{2} x = \sin x - \sin x \cos x + \cos x$$\sin^{2} x - \sin x + \sin x \cos x - \cos x = 0$$\sin x(\sin x - 1) + \cos x(\sin x - 1) = 0$$(\sin x + \cos x)(\sin x - 1) = 0$Case $1$: $\sin x = 1 \implies x = 2n \pi + \frac{\pi}{2} = n \pi + (-1)^{n} \frac{\pi}{2}$ (for $n \in \mathbb{Z}$).Case $2$: $\sin x + \cos x = 0 \implies 	an x = -1 \implies x = m \pi - \frac{\pi}{4} = m \pi + \frac{3 \pi}{4}$ (for $m \in \mathbb{Z}$).Thus,the general solution is $x = n \pi + (-1)^{n} \frac{\pi}{2}$ or $x = m \pi + \frac{3 \pi}{4}$.

The general solution of $\sin^{2} x \cdot \sec x = \tan x - \sin x + 1$ is

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