Find the general solution of 3 sin^4(theta) + | vedclass

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(D) Given equation: $3 \sin^4 	heta + \cos^4 	heta = 1$Using $\cos^4 	heta = (1 - \sin^2 	heta)^2 = 1 - 2\sin^2 	heta + \sin^4 	heta$,we get:$3

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$n\pi, n\pi + \frac{\pi}{4}, n\pi - \frac{\pi}{4}$

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(D) Given equation: $3 \sin^4 	heta + \cos^4 	heta = 1$Using $\cos^4 	heta = (1 - \sin^2 	heta)^2 = 1 - 2\sin^2 	heta + \sin^4 	heta$,we get:$3 \sin^4 	heta + (1 - 2\sin^2 	heta + \sin^4 	heta) = 1$$4 \sin^4 	heta - 2\sin^2 	heta = 0$$2 \sin^2 	heta (2 \sin^2 	heta - 1) = 0$This implies $\sin^2 	heta = 0$ or $\sin^2 	heta = \frac{1}{2}$.Case $1$: $\sin^2 	heta = 0$ $\Rightarrow \sin 	heta = 0$ $\Rightarrow 	heta = n\pi$.Case $2$: $\sin^2 	heta = \frac{1}{2} = \sin^2(\frac{\pi}{4}) \Rightarrow 	heta = n\pi \pm \frac{\pi}{4}$.Combining both cases,the general solution is $	heta = n\pi, n\pi \pm \frac{\pi}{4}$ where $n \in \mathbb{Z}$.

Find the general solution of $3 \sin^4(\theta) + \cos^4(\theta) = 1$.

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