Question

If \[3\cos x \ne 2\sin x\], then the general solution of \[{\sin ^2}x - \cos 2x = 2 - \sin 2x\] is
A. \[n\pi + {\left( { - 1} \right)^n}\dfrac{\pi }{2},n \in Z\]
B. \[\dfrac{{n\pi }}{2},n \in Z\]
C. \[\left( {4n \pm 1} \right)\dfrac{\pi }{2},n \in Z\]
D. \[\left( {2n - 1} \right),n \in Z\]

Answer 1

Hint: We are given a problem in which we need to find the general solution to the given equation. There are several approaches to solving a trigonometric equation. We need to remember the basic trigonometric equations and identities(which are mentioned below) to get the desired result.

Formula used:
1. \[{\sin ^2}x = 1 - {\cos ^2}x\]
2. \[\cos 2x = 2{\cos ^2}x - 1\]
3. \[\sin 2x = 2\sin x\cos x\]
4. \[\cos \,x = 2n\pi \pm \dfrac{\pi }{2}\]

Complete step-by-step solution:
We are given that \[{\sin ^2}x - \cos 2x = 2 - \sin 2x\]
Now in order to solve the given trigonometric equation, we should first take all the terms to the left side of the equation
 \[{\sin ^2}x - \cos 2x - 2 + \sin 2x = 0...\left( 1 \right)\]
Now we apply the formula \[{\sin ^2}x = 1 - {\cos ^2}x\] and \[\cos 2x = 2{\cos ^2}x - 1\] in equation \[\left( 1 \right)\], we get
\[\left( {1 - {{\cos }^2}x} \right) - \left( {2{{\cos }^2}x - 1} \right) - 2 + \sin 2x = 0\]
Now we simplify the above equation
\[
  1 - {\cos ^2}x - 2{\cos ^2}x + 1 - 2 + \sin 2x = 0 \\
  1 - 3{\cos ^2}x - 1 + \sin 2x = 0 \\
   - 3{\cos ^2}x + \sin 2x = 0 \\
   - 3{\cos ^2}x + 2\sin x\cos x = 0
 \]
Further solving we get,
\[\cos x( - 3\cos x + 2\sin x) = 0\]
Now it is given that \[ - 3\cos x + 2\sin x \ne 0\]
So, \[\cos x = 0\]
Now a simplified form of the trigonometric equation is represented by the above equation at odd multiples of a number, we know that cosine equals zero at \[\dfrac{\pi }{2}\]
Now the general solution of \[\cos \,x\] is \[\cos \,x = 2n\pi \pm \dfrac{\pi }{2}\]
\[
  x = 2n\pi \pm \dfrac{\pi }{2},n \in Z \\
   = \pi \left( {2n \pm \dfrac{1}{2}} \right),n \in Z \\
   = \pi \left( {\dfrac{{4n \pm 1}}{2}} \right),n \in Z \\
   = \left( {4n \pm 1} \right)\dfrac{\pi }{2},n \in Z
 \]
Therefore, the general solution of \[{\sin ^2}x - \cos 2x = 2 - \sin 2x\] is \[\left( {4n \pm 1} \right)\dfrac{\pi }{2},n \in Z\]
Hence, option (C) is correct

Additional information: A general solution is a differential equation solution that contains arbitrary constants equal to the order of the differential equation. A particular solution is one that is obtained by assigning some values to arbitrary constants from a general solution.

Note: Students should remember the formula of \[\cos 2x\],\[{\sin ^2}x\], and \[\sin 2x\] to find the exact general solution of the given equation because if he/she applies the wrong formula then it will take a long time to simplify.