Question

The general solution of $ \sin x - 3\sin 2x + \sin 3x = \cos x - 3\cos 2x + \cos 3x $ is
A. $ n\pi + \dfrac{\pi }{8} $
B. $ \dfrac{{n\pi }}{2} + \dfrac{\pi }{8} $
C. $ {( - 1)^n}\dfrac{{n\pi }}{2} + \dfrac{\pi }{8} $
D. $ 2n\pi + {\operatorname{Cos} ^{ - 1}}\dfrac{3}{2} $

Answer 1

Hint: By using the formula for finding $ \operatorname{Sin} C + \operatorname{Sin} D $ and $ \operatorname{Cos} C + \operatorname{Cos} D $ we can simplify the given equation and with knowledge of the general solution of trigonometric functions like sinx, cosx, tanx we can find out the correct answer to this question.

Complete step-by-step answer:
We are given that,
 $ \sin x - 3\sin 2x + \sin 3x = \cos x - 3\cos 2x + \cos 3x $
We can rewrite the above equation as,
 $ (\sin x + \sin 3x) - 3\sin 2x = (\cos x + \cos 3x) - 3\cos 2x $ ….(1)
Using the identities
$ \operatorname{Sin} C + \operatorname{Sin} D = 2\operatorname{Sin} \dfrac{{C + D}}{2}\operatorname{Cos} \dfrac{{C - D}}{2} $
And
 $ \operatorname{Cos} C + \operatorname{Cos} D = 2\cos \dfrac{{C + D}}{2}\cos \dfrac{{C - D}}{2} $
in equation (1), we get
 $
 \Rightarrow (2\sin \dfrac{{x + 3x}}{2}\cos \dfrac{{x - 3x}}{2}) - 3\sin 2x = (2\cos \dfrac{{x + 3x}}{2}\cos \dfrac{{x - 3x}}{2}) - 3\cos 2x \\
 \Rightarrow 2\sin \dfrac{{4x}}{2}\cos \dfrac{{ - 2x}}{2} - 3\sin 2x = 2\cos \dfrac{{4x}}{2}\cos \dfrac{{ - 2x}}{2} - 3\cos 2x \\
  2\sin 2x\cos ( - x) - 3\sin 2x = 2\cos 2x\cos ( - x) - 3\cos 2x \;
  $
Now, $ \cos x $ has a positive value when x lies in the fourth quadrant and an angle of –x signifies an angle of (360-x)° i.e. the angle lies in the fourth quadrant.
 $
  \cos (360 - x) = \cos x \\
   \Rightarrow \cos ( - x) = \cos x \;
  $
Using the above results in the equation obtained,
 $ 2\sin 2x\cos x - 3\sin 2x = 2\cos 2x\cos x - 3\cos 2x $
Taking $ \sin 2x $ common from the left-hand side and $ \cos 2x $ common from the right-hand side, we get
 $
  \sin 2x(2\cos x - 3) = \cos 2x(2\cos x - 3) \\
  \dfrac{{\sin 2x}}{{\cos 2x}} = \dfrac{{2\cos x - 3}}{{2\cos x - 3}} \\
  \tan 2x = 1 \;
  $
Now, we know that the general solution of $ \tan x = 1 $ is $ x = n\pi + \dfrac{\pi }{4} $
 $
  \therefore 2x = n\pi + \dfrac{\pi }{4} \\
  x = \dfrac{{n\pi }}{2} + \dfrac{\pi }{8} \;
  $
Hence, option (B) is the correct answer.
So, the correct answer is “Option B”.

Additional information:
 $ \sin x $ has a negative value when x lies in the fourth quadrant,
 $
  \sin (360 - x) = - \sin x \\
   \Rightarrow \sin ( - x) = - \sin x \;
  $
 $ \tan x $ also has a negative value when x lies in the fourth quadrant,
 $
  \tan (360 - x) = - \tan x \\
   \Rightarrow \tan ( - x) = - \tan x \;
  $

Note:
$
\operatorname{Sin} C + \operatorname{Sin} D = 2\operatorname{Sin} \dfrac{{C + D}}{2}\operatorname{Cos} \dfrac{{C - D}}{2} \\
  \operatorname{Sin} C - \operatorname{Sin} D = 2\operatorname{Cos} \dfrac{{C + D}}{2}\operatorname{Sin} \dfrac{{C - D}}{2} \\
  \operatorname{Cos} C + \operatorname{Cos} D = 2\operatorname{Cos} \dfrac{{C + D}}{2}\operatorname{Cos} \dfrac{{C - D}}{2} \\
  \operatorname{Cos} C - \operatorname{Cos} D = 2\operatorname{Sin} \dfrac{{C + D}}{2}\operatorname{Sin} \dfrac{{C - D}}{2} \\
 $
Remember these formulas and don’t get confused between them.
 $ \cos x $ is an even function as $ \cos ( - x) = \cos x $ ,while $ \sin x $ and $ \tan x $ are odd functions.