Mathematics - Trigonometric Equation Question with Solution | TestHub

We use the identity ${\sin }^{6}x+{\cos }^{6}x=({\sin }^{2}x+{\cos }^{2}x)({\sin }^{4}x-{\sin }^{2}x{\cos }^{2}x+{\cos }^{4}x)=1\cdot (({\sin }^{2}x+{\cos }^{2}x{)}^2-3{\sin }^{2}x{\cos }^{2}x)=1-3{\sin }^{2}x{\cos }^{2}x$. Substituting this into the equation, we get $1-3{\sin }^{2}x{\cos }^{2}x=\sin x\cos x$. Let $y=\sin x\cos x$. The equation becomes $1-3y^2=y$, which rearranges to $3y^2+y-1=0$. Using the quadratic formula, $y=\frac{-1\pm \sqrt{1^2-4(3)(-1)}}{2(3)}=\frac{-1\pm \sqrt{13}}{6}$. Since $y=\sin x\cos x=\frac{1}{2}\sin 2x$, we know that $y$ must be in the range $[-\frac{1}{2},\frac{1}{2}]$. The value $y_1=\frac{-1+\sqrt{13}}{6}$ is approximately $\frac{-1+3.6}{6}\approx 0.43$, which is in $[-\frac{1}{2},\frac{1}{2}]$. The value $y_2=\frac{-1-\sqrt{13}}{6}$ is approximately $\frac{-1-3.6}{6}\approx -0.76$, which is outside the range. Thus, we only consider $\sin x\cos x=\frac{-1+\sqrt{13}}{6}$. This implies $\frac{1}{2}\sin 2x=\frac{-1+\sqrt{13}}{6}$, so $\sin 2x=\frac{-1+\sqrt{13}}{3}$. Let $k=\frac{-1+\sqrt{13}}{3}$. Since $3<\sqrt{13}<4$, we have $\frac{2}{3}<k<1$. Thus, $k$ is a valid value for $\sin 2x$. Let $\alpha =\arcsin (k)$, where $0<\alpha <\frac{\pi }{2}$. The general solution for $\sin 2x=k$ is $2x=m\pi +(-1{)}^m\alpha$, for integer $m$. We need solutions in $[0,3\pi ]$, so $2x\in [0,6\pi ]$. For $m=0,2x=\alpha ⟹x=\frac{\alpha }{2}$. For $m=1,2x=\pi -\alpha ⟹x=\frac{\pi -\alpha }{2}$. For $m=2,2x=2\pi +\alpha ⟹x=\pi +\frac{\alpha }{2}$. For $m=3,2x=3\pi -\alpha ⟹x=\frac{3\pi -\alpha }{2}$. For $m=4,2x=4\pi +\alpha ⟹x=2\pi +\frac{\alpha }{2}$. For $m=5,2x=5\pi -\alpha ⟹x=\frac{5\pi -\alpha }{2}$. For $m=6,2x=6\pi +\alpha ⟹x=3\pi +\frac{\alpha }{2}$, which is outside the interval $[0,3\pi ]$. All 6 values listed are distinct and lie within the interval $[0,3\pi ]$. Therefore, there are 6 solutions.