Question

Find all angles, 0º < 0 < 360°, that satisfy the equation below, to the nearest 10th

of a degree.
2 tan^2(theta)+ tan(theta)= 0

Answer 1

`2tan^2(\theta )+tan(\theta )=0\implies tan(\theta )[2tan(\theta )+1)]=0 \\\\[-0.35em] ~\dotfill\\\\ tan(\theta)=0\implies \theta=tan^(-1)(0)\implies \theta= \begin{cases} 0\\ 360^o \end{cases} \\\\[-0.35em] ~\dotfill\\\\ 2tan(\theta)+1=0\implies 2tan(\theta)=-1\implies tan(\theta)=-\cfrac{1}{2} \\\\\\ \theta=tan^(-1)\left( -\cfrac{1}{2} \right)\implies \theta \approx \begin{cases} \stackrel{180^o~~ - ~~26.57}{153.43^o}\\ \stackrel{360~~ - ~~26.57}{333.43^o} \end{cases}`

for the second pair of angles, the angle of reference is really about 26.57°. Now, the tangent is only negative in two quadrants, the II and IV Quadrants, using the angle of reference, it'd give us those angles.

Make sure your calculator is in Degree mode.