## Trigonometry (11th Edition) Clone

The solution set to this problem is $$\{53.6^\circ, 126.4^\circ, 187.9^\circ, 352.1^\circ\}$$
$$9\sin^2\theta-6\sin\theta=1$$ over interval $[0^\circ,360^\circ)$ 1) Solve the equation: $$9\sin^2\theta-6\sin\theta=1$$ $$9\sin^2\theta-6\sin\theta-1=0$$ Here we need to consider the equation a quadratic formula. In fact, if we take $\sin\theta$ as $x$, we would have $$9x^2-6x-1=0$$ with $a=9, b=6$ and $c=-1$ - Calculate $\Delta$: $\Delta=b^2-4\times a\times c=(-6)^2-4\times9\times(-1)=36-(-36)=72$ - Calculate $x$, or in other words, $\sin\theta$: $$x=\sin\theta=\frac{-b\pm\sqrt\Delta}{2a}=\frac{6\pm\sqrt{72}}{18}=\frac{6\pm6\sqrt2}{18}=\frac{1\pm\sqrt2}{3}$$ 2) Apply the inverse function: - For $\sin\theta=\frac{1+\sqrt2}{3}$: $$\sin\theta=\frac{1+\sqrt2}{3}\approx0.805$$ which means $$\theta=\sin^{-1}0.805$$ $$\theta\approx53.6^\circ\hspace{1cm}\text{or}\hspace{1cm}\theta\approx126.4^\circ$$ - For $\sin\theta=\frac{1-\sqrt2}{3}$: $$\sin\theta=\frac{1-\sqrt2}{3}\approx-0.138$$ which means $$\theta=\sin^{-1}(-0.138)$$ $$\theta\approx187.9^\circ\hspace{1cm}\text{or}\hspace{1cm}\theta\approx352.1^\circ$$ Therefore, $$\theta\in\{53.6^\circ, 126.4^\circ, 187.9^\circ, 352.1^\circ\}$$ In other words, the solution set to this problem is $$\{53.6^\circ, 126.4^\circ, 187.9^\circ, 352.1^\circ\}$$