The solutions are \displaystyle{S}={\left\lbrace{60}^{\circ},{300}^{\circ}\right\rbrace} for \displaystyle\theta\in{\left[{0},{360}\right)} Explanation: This is a quadratic equation in \displaystyle{\cos{\theta}} ...

The solutions are \displaystyle{S}={\left\lbrace{0},\frac{\pi}{{3}},\frac{{{5}\pi}}{{3}}\right\rbrace} Explanation: Let's factorise the LHS \displaystyle{2}{{\cos}^{{2}}\theta}-{3}{\cos{\theta}}+{1}={\left({2}{\cos{\theta}}-{1}\right)}{\left({\cos{\theta}}-{1}\right)}={0} ...

\displaystyle\theta={150.56} deg Explanation: \displaystyle{2}{{\cos}^{{2}}\theta}-{4}{\cos{\theta}}-{5}={0} . Note that this equation looks like the quadratic equation \displaystyle{2}{x}^{{2}}-{4}{x}-{5}={0} ...

\displaystyle{\left({1}+{2}{k}\right)}\pi;\frac{\pi}{{3}}+{2}{k}\pi;\frac{{{5}\pi}}{{3}}+{2}{k}\pi Explanation: Solve the quadratic equation for cos t. \displaystyle{f{{\left({t}\right)}}}={2}{{\cos}^{{2}}{t}}+{\cos{{t}}}-{1}={0} ...

See below. Explanation: At point \displaystyle{x}_{{0}},{y}_{{0}} there is a tangent line and a normal line given by \displaystyle{L}_{{t}}\to{y}={y}_{{0}}+{m}_{{0}}{\left({x}-{x}_{{0}}\right)} ...

Factor out a cos. Explanation: \displaystyle{\cos{\theta}}{\left({2}{\cos{\theta}}+{1}\right)}={0} \displaystyle{\cos{\theta}}={0}{\quad\text{or}\quad}{2}{\cos{\theta}}+{1}={0} \displaystyle\theta=\pi,\frac{{{3}\pi}}{{2}}{\quad\text{or}\quad}{\cos{\theta}}=-\frac{{1}}{{2}} ...