The equation 3sin^2 x + 10cos x - 6 = 0 is sa | vedclass

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(B) Given equation: $3\sin^2 x + 10\cos x - 6 = 0$Using the identity $\sin^2 x = 1 - \cos^2 x$,we get:$3(1 - \cos^2 x) + 10\cos x - 6 = 0$$3 - 3\cos^2

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$x = 2n\pi \pm \cos^{-1}(1/3)$

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(B) Given equation: $3\sin^2 x + 10\cos x - 6 = 0$Using the identity $\sin^2 x = 1 - \cos^2 x$,we get:$3(1 - \cos^2 x) + 10\cos x - 6 = 0$$3 - 3\cos^2 x + 10\cos x - 6 = 0$$-3\cos^2 x + 10\cos x - 3 = 0$$3\cos^2 x - 10\cos x + 3 = 0$Factoring the quadratic equation:$3\cos^2 x - 9\cos x - \cos x + 3 = 0$$3\cos x(\cos x - 3) - 1(\cos x - 3) = 0$$(3\cos x - 1)(\cos x - 3) = 0$This gives two possibilities:$1) \cos x = 3$ (Not possible,as $-1 \le \cos x \le 1$)$2) \cos x = 1/3$The general solution for $\cos x = \cos \alpha$ is $x = 2n\pi \pm \alpha$.Therefore,$x = 2n\pi \pm \cos^{-1}(1/3)$.

The equation $3\sin^2 x + 10\cos x - 6 = 0$ is satisfied,if

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