## Trigonometry 7th Edition

$\theta=60^{\circ}+180^{\circ}k,90^{\circ}+180^{\circ}k,120^{\circ}+180^{\circ}k$ where k is any integer.
$2(\cos2\theta)^{2}+3\cos2\theta+1=0$ or, $(2cos2\theta+1)(\cos2\theta+1)=0$ either $cos2\theta=-\frac{1}{2}$ or $-1$ For, $\cos2\theta=-\frac{1}{2}$, $2\theta=\cos^{-1}(-\frac{1}{2})$, Since, $\cos2\theta$ is negative in both quadrants 2 and 3, $2\theta=180^{\circ}-60^{\circ}=120^{\circ} \,\,or \,\,180^{\circ}+60^{\circ}=240^{\circ},$ To find all values of $2\theta$, we can add $360^{\circ}k$ where k is any integer, since the period of cos function is $360^{\circ}$. So, $2\theta=120^{\circ}+360^{\circ}k\,\,or \,\,240^{\circ}+360^{\circ}k,$ So, $\theta=60^{\circ}+180^{\circ}k\,\,or \,\,120^{\circ}+180^{\circ}k,$ For, $\cos2\theta=-1$, $2\theta=\cos^{-1}(-1)$, $2\theta=180^{\circ}$ To find all values of $2\theta$, we can add $360^{\circ}k$ where k is any integer, since the period of cos function is $360^{\circ}$. So, $2\theta=180^{\circ}+360^{\circ}k\,\,$ So, $\theta=90^{\circ}+180^{\circ}k\,\,$ So, all degree solutions are $\theta=60^{\circ}+180^{\circ}k,90^{\circ}+180^{\circ}k,120^{\circ}+180^{\circ}k$