Question 25

Maths Question: 

$\begin{align}  & \text{Obtain the first general solution of the  equation and then use the general } \\ & \text{solution to find all solution in the range }{{0}^{\circ }}\le \theta \le {{360}^{\circ }} \\ & 7{{\sec }^{2}}\theta =6\tan \theta +8 \\\end{align}$

Maths Solution: 

$\begin{align}  & 7{{\sec }^{2}}\theta =6\tan \theta +8 \\ & 7(1+{{\tan }^{2}}\theta )=6\tan \theta +8 \\ & 7{{\tan }^{2}}\theta -6\tan \theta -1=0 \\ & 7{{\tan }^{2}}\theta -7\tan \theta +\tan \theta -1=0 \\ & 7\tan \theta (\tan \theta -1)+1(\tan \theta -1)=0 \\ & (7\tan \theta +1)(\tan \theta -1)=0 \\ & \tan \theta =-\frac{1}{7}\text{ or }\tan \theta =1 \\ & \text{When }\tan \theta =-\frac{1}{7},\text{ }\theta =-{{8}^{\circ }}8' \\ & \text{The general solution }\theta =n{{180}^{\circ }}+\alpha  \\ & \theta =n{{180}^{\circ }}+(-{{8}^{\circ }}8') \\ & \theta ={{171}^{\circ }}52',\text{ }{{351}^{\circ }}21' \\ & \text{When }\tan \theta =1 \\ & \theta ={{\tan }^{-1}}(1)={{45}^{\circ }} \\ & \text{The general solution }\theta =n{{180}^{\circ }}+{{45}^{\circ }} \\ & \theta ={{45}^{\circ }},{{225}^{\circ }} \\ & \text{The values of }\theta \text{ are }{{45}^{\circ }},{{171}^{\circ }}52',{{225}^{\circ }},{{351}^{\circ }}21' \\\end{align}$

University mathstopic: