Equation of the Type : a sin θ ± b cos θ = c

If a and b are positive,                           $ \displaystyle a\sin \theta \pm b\cos \theta $ can be written in the form $ \displaystyle R\sin (\theta \pm \alpha ),$                           $ \displaystyle a\cos \theta \pm b\sin \theta $ can be written in the form $\displaystyle R\sin (\theta \mp \alpha ),$   where $ R = \sqrt{a^2 + b^2}, R \cos \alpha = a , R \sin \alpha = b$ and $ \displaystyle \tan \alpha =\frac{b}{a}$ with $ \displaystyle {{0}^{{}^\circ }}<\alpha <{{90}^{{}^\circ }}.$

Example (1)      Solve the equation $ \displaystyle 8\sin \theta +6\cos \theta =5$ for $ \displaystyle {{0}^{{}^\circ }}\le \theta \le {{360}^{{}^\circ }}$.

Solution
         
              $ \displaystyle 8\sin \theta +6\cos \theta =5$

              Let $ \displaystyle R\cos \alpha =8$ and $ \displaystyle R\sin \alpha =6$.

              $ \displaystyle \therefore R=\sqrt{{{{8}^{2}}+{{6}^{2}}}}=\sqrt{{100}}=10$ and $ \displaystyle \tan \alpha =\frac{6}{8}\Rightarrow \alpha =36{}^\circ 5{2}'$

              Since $ \displaystyle 8\sin \theta +6\cos \theta =R\sin (\theta +\alpha )$,

             $ \displaystyle R\sin (\theta +\alpha )=5\Rightarrow 10\sin (\theta +36{}^\circ 5{2}')=5\Rightarrow \sin (\theta +36{}^\circ 5{2}')=\frac{1}{2}$

             $ \displaystyle \begin{array}{l}\therefore \theta +36{}^\circ 5{2}'=30{}^\circ \\\end{array}$ (1^st quadrant) or

                 $ \displaystyle \theta +36{}^\circ 5{2}'=150{}^\circ $(2^nd quadrant) or

                 $ \displaystyle \theta +36{}^\circ 5{2}'=390{}^\circ $(1^st quadrant)

             $ \displaystyle \therefore \theta =-6{}^\circ 5{2}'$ or $ \displaystyle \ \theta =113{}^\circ {8}'$ or $ \displaystyle \theta =353{}^\circ {8}'$

             Since $ \displaystyle {{0}^{{}^\circ }}\le \theta \le {{360}^{{}^\circ }}$, $ \displaystyle \theta =-6{}^\circ 5{2}'\ $is impossible.
             
             $ \displaystyle \therefore \theta =113{}^\circ {8}'$ or $ \displaystyle \theta =353{}^\circ {8}'$.