## Trigonometry (11th Edition) Clone

There are 2 values of $\theta$ in need to find, which is {$57.7^\circ, 159.2^\circ$}.
$$\cot\theta+2\csc\theta=3$$ $$2\csc\theta=3-\cot\theta$$ Remember the identity $\csc^2\theta=1+\cot^2\theta$, which can be used to change the equation into a single trigonometric equation. So what we do is square each side and express $\csc^2\theta$ in terms of $cot^2\theta$, like this $$4\csc^2\theta=(3-\cot\theta)^2$$ $$4(1+\cot^2\theta)=9+\cot^2\theta-6\cot\theta$$ $$4+4\cot^2\theta=9+\cot^2\theta-6\cot\theta$$ $$3\cot^2\theta+6\cot\theta-5=0$$ Now we solve the equation like a normal quadratic equation, with $a=3, b=6, c=-5$. $$\cot\theta=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ $$\cot\theta=\frac{-6\pm\sqrt{6^2-4\times3\times(-5)}}{2\times3}$$ $$\cot\theta=\frac{-6\pm\sqrt{36+60}}{6}$$ $$\cot\theta=\frac{-6\pm\sqrt{96}}{6}$$ $$\cot\theta=\frac{-6\pm2\sqrt{24}}{6}$$ $$\cot\theta=\frac{-3\pm\sqrt{24}}{3}$$ $$\cot\theta\approx0.633$$ or $$\cot\theta\approx-2.633$$ *For $\cot\theta=0.633$, $\theta=\cot^{-1}(0.633)=\tan^{-1}(\frac{1}{0.633})$ $\theta$ would have 2 values {$57.7^\circ,237.7^\circ$} Now we must do the checking. - With $\theta=57.7^\circ$, $\cot(57.7^\circ)+2\csc(57.7^\circ)=0.632+2\times1.183\approx2.998\approx3$ (right) - With $\theta=237.7^\circ$, $\cot(237.7^\circ)+2\csc(237.7^\circ)=0.632+2\times(-1.183)\approx-1.734\ne3$ (wrong) *For $\cot\theta=-2.633$, $\theta=\cot^{-1}(-2.633)=\tan^{-1}(\frac{-1}{2.633})$ $\theta$ would have 2 values {$159.2^\circ,339.2^\circ$} Now we must do the checking. - With $\theta=159.2^\circ$, $\cot(159.2^\circ)+2\csc(159.2^\circ)=-2.633+2\times2.816\approx2.999\approx3$ (right) - With $\theta=339.2^\circ$, $\cot(339.2^\circ)+2\csc(339.2^\circ)=-2.633+2\times(-2.816)\approx-8.265\ne3$ (wrong) Overall, there are 2 values of $\theta$ in need to find, which is {$57.7^\circ, 159.2^\circ$}.