Question 27

Maths Question: 

$\begin{align}  & \text{Obtain the first general solution of the  equation and then use the general } \\ & \text{solution to find all solution in the range }{{0}^{\circ }}\le \theta \le {{360}^{\circ }} \\ & 2{{\cos }^{2}}\theta =\sin \theta +1 \\\end{align}$

Maths Solution: 

$\begin{align}  & 2{{\cos }^{2}}\theta =\sin \theta +1 \\ & 2{{\cos }^{2}}\theta -\sin \theta -1=0 \\ & 2(1-{{\sin }^{2}}\theta )-\sin \theta -1=0 \\ & 2-2{{\sin }^{2}}\theta -\sin \theta -1=0 \\ & 2{{\sin }^{2}}\theta +\sin \theta -1=0 \\ & (2\sin \theta -1)(\sin \theta +1)=0 \\ & \sin \theta =\tfrac{1}{2},\text{ }\sin \theta =-1 \\ & \text{when }\sin \theta =\tfrac{1}{2},\text{  }\theta ={{30}^{\circ }} \\ & \text{General solution }\theta =n{{180}^{\circ }}+{{(-1)}^{n}}({{30}^{\circ }}) \\ & \theta ={{30}^{\circ }}\text{,15}{{\text{0}}^{\circ }} \\ & \text{when }\sin \theta =-1,\text{ }\theta ={{\sin }^{-1}}(-1)=-{{90}^{^{\circ }}} \\ & \text{General solution }\theta =n{{180}^{^{\circ }}}+(-1)(-{{90}^{\circ }}) \\ & \theta ={{270}^{\circ }} \\ & \text{The value of }\theta \,\text{are }{{30}^{\circ }},{{150}^{\circ }},{{270}^{\circ }} \\\end{align}$

University mathstopic: