Question

Find a general solution for $ x $ if $ \sin 8x - \cos 6x = \sqrt 3 \left( {\sin 6x + \cos 8x} \right) $ .

Answer 1

Hint: In this problem, first we will simplify the given equation. Then, we will use the values of trigonometric functions for particular angles. Then, we will use some trigonometric identities and formulas to find the general solution of the given problem.

Complete step-by-step answer:
In the given problem, we have to find $ x $ if $ \sin 8x - \cos 6x = \sqrt 3 \left( {\sin 6x + \cos 8x} \right) \cdots \cdots \left( 1 \right) $ . Let us simplify the equation $ \left( 1 \right) $ . So, we can write
 $
  \sin 8x - \cos 6x = \sqrt 3 \sin 6x + \sqrt 3 \cos 8x \\
   \Rightarrow \sqrt 3 \cos 8x - \sin 8x = - \cos 6x - \sqrt 3 \sin 6x \\
   \Rightarrow \sqrt 3 \cos 8x - \sin 8x = - \left( {\cos 6x + \sqrt 3 \sin 6x} \right) \cdots \cdots \left( 2 \right) \\
  $
Let us divide by $ 2 $ to each term on both sides of equation $ \left( 2 \right) $ . So, we can write
 $ \dfrac{{\sqrt 3 }}{2}\cos 8x - \dfrac{1}{2}\sin 8x = - \left( {\dfrac{1}{2}\cos 6x + \dfrac{{\sqrt 3 }}{2}\sin 6x} \right) \cdots \cdots \left( 3 \right) $
We know that $ \cos \dfrac{\pi }{6} = \dfrac{{\sqrt 3 }}{2} $ and $ \sin \dfrac{\pi }{6} = \dfrac{1}{2} $ . Use this information on the LHS of equation $ \left( 3 \right) $ . Also we know that $ \cos \dfrac{\pi }{3} = \dfrac{1}{2} $ and $ \sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2} $ . Use this information on the RHS of the equation $ \left( 3 \right) $ . So, we can write
 $ \cos \left( {\dfrac{\pi }{6}} \right)\cos 8x - \sin \left( {\dfrac{\pi }{6}} \right)\sin 8x = - \left[ {\cos \left( {\dfrac{\pi }{3}} \right)\cos 6x + \sin \left( {\dfrac{\pi }{3}} \right)\sin 6x} \right] \cdots \cdots \left( 4 \right) $
We know that $ \cos A\cos B - \sin A\sin B = \cos \left( {A + B} \right) $ . Use this information on LHS of the equation $ \left( 4 \right) $ . Also we know that $ \cos A\cos B + \sin A\sin B = \cos \left( {A - B} \right) $ . Use this information on the RHS of equation $ \left( 4 \right) $ . So, we can write
 $ \cos \left( {8x + \dfrac{\pi }{6}} \right) = - \cos \left( {6x - \dfrac{\pi }{3}} \right) \cdots \cdots \left( 5 \right) $
We know that $ \cos \left( {\pi - \theta } \right) = - \cos \theta $ . Use this information on the RHS of the equation $ \left( 5 \right) $ . So, we can write
 $
  \cos \left( {8x + \dfrac{\pi }{6}} \right) = \cos \left[ {\pi - \left( {6x - \dfrac{\pi }{3}} \right)} \right] \\
   \Rightarrow \cos \left( {8x + \dfrac{\pi }{6}} \right) = \cos \left( { - 6x + \dfrac{{4\pi }}{3}} \right) \cdots \cdots \left( 6 \right) \\
  $
We know that for any real numbers $ x $ and $ y $ , $ \cos x = \cos y \Rightarrow x = 2n\pi \pm y $ where $ n $ is integer. Use this information in equation $ \left( 6 \right) $ . So, we can write
 $ 8x + \dfrac{\pi }{6} = 2n\pi \pm \left( { - 6x + \dfrac{{4\pi }}{3}} \right) \cdots \cdots \left( 7 \right) $ where $ n $ is integer.
Let us solve the equation $ \left( 7 \right) $ for $ x $ by considering positive signs on RHS. So, we can write
 $
  8x + \dfrac{\pi }{6} = 2n\pi + \left( { - 6x + \dfrac{{4\pi }}{3}} \right) \\
   \Rightarrow 8x + 6x = 2n\pi + \dfrac{{4\pi }}{3} - \dfrac{\pi }{6} \\
   \Rightarrow 14x = 2n\pi + \dfrac{{7\pi }}{6} \\
   \Rightarrow x = \dfrac{{n\pi }}{7} + \dfrac{\pi }{{12}} \\
  $
Let us solve the equation $ \left( 7 \right) $ for $ x $ by considering negative sign on RHS. So, we can write
 $
  8x + \dfrac{\pi }{6} = 2n\pi - \left( { - 6x + \dfrac{{4\pi }}{3}} \right) \\
   \Rightarrow 8x - 6x = 2n\pi - \dfrac{{4\pi }}{3} - \dfrac{\pi }{6} \\
   \Rightarrow 2x = 2n\pi - \dfrac{{3\pi }}{2} \\
   \Rightarrow x = n\pi - \dfrac{{3\pi }}{4} \\
  $
Hence, $ x = \dfrac{{n\pi }}{7} + \dfrac{\pi }{{12}} $ or $ x = n\pi - \dfrac{{3\pi }}{4} $ if $ \sin 8x - \cos 6x = \sqrt 3 \left( {\sin 6x + \cos 8x} \right) $ . This is the general solution of the given problem.

Note: In this type of problems, trigonometric identities and formulas are very useful to find the general solution. Remember that $ \sin A\cos B + \cos A\sin B = \sin \left( {A + B} \right) $ and $ \sin A\cos B - \cos A\sin B = \sin \left( {A - B} \right) $ . Also remember that for any real numbers $ x $ and $ y $ , $ \sin x = \sin y \Rightarrow x = n\pi + {\left( { - 1} \right)^n}y $ where $ n $ is integer.