<p>Total number of solutions of \(\sin^4 x + \cos^4 x = \sin x \times \cos x\) in \([0, 2\pi]\) is equal to</p>
Step-by-Step Solution
Key Concept: Use the identity $\sin^4 x + \cos^4 x = 1 - 2\sin^2 x \cos^2 x$ and substitute $u = \sin x \cos x$ to convert to a quadratic equation.
<p>Simplify using $\sin^4 x + \cos^4 x = 1 - 2\sin^2 x \cos^2 x$. The equation becomes $1 - 2\sin^2 x \cos^2 x = \sin x \cos x$. Let $u = \sin x \cos x = \frac{\sin 2x}{2}$, giving $1 - 2u^2 = u$, or $2u^2 + u - 1 = 0$. Solving yields 8 solutions in $[0, 2\pi]$.</p>
Correct Answer: d