Question

Find the general solution of the equation $\sin^4 x + \cos^4 x = \sin x \cos x$

• A. n\pi + \frac{\pi}{4}
• B. n\pi - \frac{\pi}{4}
• C. 2n\pi + \frac{\pi}{4}
• D. 2n\pi - \frac{\pi}{4}

Answer 1

Answer: (A)

n\pi + \frac{\pi}{4}

Answer 2

The equation can be rewritten as $1 - 2\sin^2 x \cos^2 x = \sin x \cos x$. Using $\sin x \cos x = \frac{1}{2}\sin(2x)$ and $\sin^2 x \cos^2 x = \frac{1}{4}\sin^2(2x)$, we get $1 - \frac{1}{2}\sin^2(2x) = \frac{1}{2}\sin(2x)$. This simplifies to $\sin^2(2x) + \sin(2x) - 2 = 0$, which factors as $(\sin(2x)+2)(\sin(2x)-1)=0$. Since $\sin(2x) \neq -2$, we have $\sin(2x) = 1$. The general solution is $2x = 2n\pi + \frac{\pi}{2}$, thus $x = n\pi + \frac{\pi}{4}$.