Let S be the sum of all solutions (in radians) of the equation ^4+^4 - = 0 in [0,4].

Question

Let $S$ be the sum of all solutions (in radians) of the equation $\sin^4	heta+\cos^4	heta - \sin	heta\cos	heta = 0$ in $[0,4\pi]$. Then $\dfrac{8S}\pi$ is equal to:

Answer: 56

Accepted Answer

$$1-2\sin^2\theta\cos^2\theta - \sin\theta\cos\theta = 0 \implies 2-\sin^22\theta-\sin2\theta=0 \implies \sin^22\theta+\sin2\theta-2=0$$

$$(\sin2\theta+2)(\sin2\theta-1)=0 \implies \sin2\theta=1$$

Within $2\theta\in[0,8\pi]$: $2\theta = \tfrac\pi2,\tfrac{5\pi}2,\tfrac{9\pi}2,\tfrac{13\pi}2$, so $\theta = \tfrac\pi4,\tfrac{5\pi}4,\tfrac{9\pi}4,\tfrac{13\pi}4$. Sum $S = \dfrac{28\pi}4 = 7\pi$.

$$\frac{8S}\pi = \frac{8\times7\pi}\pi = 56$$

Solutions · 1