If the equation sin 3theta - cos^2 theta = fr | vedclass

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(C) Given equation: $\sin 3	heta - \cos^2 	heta = \frac{1}{4}$.Using $\sin 3	heta = 3\sin 	heta - 4\sin^3 	heta$ and $\cos^2 	heta = 1 - \sin^2

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$4$

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(C) Given equation: $\sin 3	heta - \cos^2 	heta = \frac{1}{4}$.Using $\sin 3	heta = 3\sin 	heta - 4\sin^3 	heta$ and $\cos^2 	heta = 1 - \sin^2 	heta$, we get:$3\sin 	heta - 4\sin^3 	heta - (1 - \sin^2 	heta) = \frac{1}{4}$.$-4\sin^3 	heta + \sin^2 	heta + 3\sin 	heta - 1 = \frac{1}{4}$.Multiply by $-4$: $16\sin^3 	heta - 4\sin^2 	heta - 12\sin 	heta + 4 = -1$.$16\sin^3 	heta - 4\sin^2 	heta - 12\sin 	heta + 5 = 0$.Let $x = \sin 	heta$. Since $	heta \in [0, \pi]$, $x \in [0, 1]$.$f(x) = 16x^3 - 4x^2 - 12x + 5 = 0$.$f(0) = 5$ and $f(1) = 16 - 4 - 12 + 5 = 5$.$f'(x) = 48x^2 - 8x - 12 = 4(12x^2 - 2x - 3)$.Roots of $f'(x) = 0$ are $x = \frac{2 \pm \sqrt{4 + 144}}{24} = \frac{2 \pm \sqrt{148}}{24} = \frac{1 \pm \sqrt{37}}{12}$.Since $x \in [0, 1]$, consider $x_0 = \frac{1 + \sqrt{37}}{12} \approx \frac{1 + 6.08}{12} \approx 0.59$.$f(0.59) = 16(0.59)^3 - 4(0.59)^2 - 12(0.59) + 5 \approx 3.28 - 1.39 - 7.08 + 5 = -0.19$.Since $f(0) > 0$, $f(x_0)  0$, there are two roots for $x$ in $(0, 1)$.For each $x \in (0, 1)$, there are two values of $	heta$ in $[0, \pi]$ such that $\sin 	heta = x$.Thus, there are $2 	imes 2 = 4$ solutions.

If the equation $\sin 3\theta - \cos^2 \theta = \frac{1}{4}$ and $\theta \in [0, \pi]$, then the number of solutions is...

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