## Trigonometry (11th Edition) Clone

The solution set to this problem is $$\{78.1^\circ, 281.9^\circ\}$$
$$4\cos^2\theta+4\cos\theta=1$$ over interval $[0^\circ,360^\circ)$ 1) Solve the equation: $$4\cos^2\theta+4\cos\theta=1$$ $$4\cos^2\theta+4\cos\theta-1=0$$ Here we need to consider the equation a quadratic formula. In fact, if we take $\cos\theta$ as $x$, we would have $$4x^2+4x-1=0$$ with $a=4, b=4$ and $c=-1$ - Calculate $\Delta$: $\Delta=b^2-4\times a\times c=4^2-4\times4\times(-1)=16-(-16)=32$ - Calculate $x$, or in other words, $\cos\theta$: $$x=\cos\theta=\frac{-b\pm\sqrt\Delta}{2a}=\frac{-4\pm\sqrt{32}}{8}=\frac{-4\pm4\sqrt2}{8}=\frac{-1\pm\sqrt2}{2}$$ 2) Apply the inverse function: - For $\cos\theta=\frac{-1+\sqrt2}{2}$: $$\cos\theta=\frac{-1+\sqrt2}{2}\approx0.207$$ which means $$\theta=\cos^{-1}0.207$$ $$\theta\approx78.1^\circ\hspace{1cm}\text{or}\hspace{1cm}\theta\approx281.9^\circ$$ - For $\cos\theta=\frac{-1-\sqrt2}{2}$: $$\cos\theta=\frac{-1-\sqrt2}{2}\approx-1.207\lt(-1)$$ Yet we know that the range of cosine is $[-1, 1]$. Therefore, there are no values of $\theta$ which $\cos\theta$ could equal $-1.207$. Therefore, $$\theta\in\{78.1^\circ, 281.9^\circ\}$$ In other words, the solution set to this problem is $$\{78.1^\circ, 281.9^\circ\}$$