Question

Solve for x, \[\left( -\pi \le x\le \pi \right)\], the equation: $2\left( \cos x+\cos 2x \right)+\sin 2x\left( 1+2\cos x \right)=2\sin x.$
How many distinct values of x satisfy the equation in the above range?
(a) 2
(b) 4
(c) 5
(d) 6

Answer 1

Hint: Take all the terms to the left hand side (L.H.S). Apply the half angle formula to convert $\sin 2x$ into $2\sin x\cos x$. Now, take $2\sin x$ common from the last two terms and use the conversion: $2{{\cos }^{2}}x-1=\cos 2x$ to simplify the terms. Now take all the common terms together and write the equation as a product of trigonometric terms. Equate each term equal to zero and use the formula for the general solution of the trigonometric functions. Use the relation: $\cos x=0\Rightarrow x=\left( 2n+1 \right)\dfrac{\pi }{2}$, where n is any integer. Remember that we have to choose such values of n for which, \[\left( -\pi \le x\le \pi \right)\].

Complete step-by-step answer:
We have been given: $2\left( \cos x+\cos 2x \right)+\sin 2x\left( 1+2\cos x \right)=2\sin x$. Taking all the terms to L.H.S, we have,
$2\left( \cos x+\cos 2x \right)+\sin 2x\left( 1+2\cos x \right)-2\sin x=0$
Using the formula: $\sin 2x=2\sin x\cos x$, we get,
$2\left( \cos x+\cos 2x \right)+2\sin x\cos x\left( 1+2\cos x \right)-2\sin x=0$
Taking $2\sin 2x$ common from the 2nd and 3rd term, we get,
$\begin{align}
  & 2\left( \cos x+\cos 2x \right)+2\sin x\left\{ \cos x\left( 1+2\cos x \right)-1 \right\}=0 \\
 & \Rightarrow 2\left( \cos x+\cos 2x \right)+2\sin x\left\{ \cos x+2{{\cos }^{2}}x-1 \right\}=0 \\
\end{align}$
Using the identity: $2{{\cos }^{2}}x-1=\cos 2x$, we have,
$\begin{align}
  & 2\left( \cos x+\cos 2x \right)+2\sin x\left\{ \cos x\left( 1+2\cos x \right)-1 \right\}=0 \\
 & \Rightarrow 2\left( \cos x+\cos 2x \right)+2\sin x\left\{ \cos x+\cos 2x \right\}=0 \\
\end{align}$
Dividing both sides by 2, we get,
$\begin{align}
  & \left( \cos x+\cos 2x \right)+\sin x\left\{ \cos x+\cos 2x \right\}=0 \\
 & \Rightarrow \left( \cos x+\cos 2x \right)\left( 1+\sin x \right)=0 \\
\end{align}$
Using the formula: $\cos a+\cos b=2\cos \left( \dfrac{a+b}{2} \right)\cos \left( \dfrac{a-b}{2} \right)$, we get,
$\begin{align}
  & 2\cos \left( \dfrac{x+2x}{2} \right)\cos \left( \dfrac{x-2x}{2} \right)\left( 1+\sin x \right)=0 \\
 & \Rightarrow 2\cos \left( \dfrac{3x}{2} \right)\cos \left( \dfrac{-x}{2} \right)\left( 1+\sin x \right)=0 \\
\end{align}$
We know that, $\cos \left( -a \right)=\cos a$, therefore,
$\begin{align}
  & 2\cos \left( \dfrac{3x}{2} \right)\cos \left( \dfrac{x}{2} \right)\left( 1+\sin x \right)=0 \\
 & \Rightarrow \cos \left( \dfrac{3x}{2} \right)\cos \left( \dfrac{x}{2} \right)\left( 1+\sin x \right)=0 \\
\end{align}$
Substituting each term equal to 0, we get,
(i) Consider $\cos \left( \dfrac{3x}{2} \right)=0$
Using the formula for general solution: $\cos x=0\Rightarrow x=\left( 2n+1 \right)\dfrac{\pi }{2}$, we get,
$\begin{align}
  & \dfrac{3x}{2}=\left( 2n+1 \right)\dfrac{\pi }{2} \\
 & \Rightarrow x=\left( 2n+1 \right)\dfrac{\pi }{3} \\
\end{align}$
Now, we must take such integer values of n for which \[\left( -\pi \le x\le \pi \right)\]. Therefore,
For n = 0, we have,
$x=\dfrac{\pi }{3}$
For n = -1, we have,
$\begin{align}
  & x=\left( 2\times \left( -1 \right)+1 \right)\dfrac{\pi }{3} \\
 & \Rightarrow x=\left( -2+1 \right)\dfrac{\pi }{3} \\
 & \Rightarrow x=\dfrac{-\pi }{3} \\
\end{align}$
For n = -2, we have,
$\begin{align}
  & \Rightarrow x=\left( 2\times \left( -2 \right)+1 \right)\dfrac{\pi }{3} \\
 & \Rightarrow x=\left( -4+1 \right)\dfrac{\pi }{3} \\
 & \Rightarrow x=-3\times \dfrac{\pi }{3} \\
 & \Rightarrow x=-\pi \\
\end{align}$
Now, there are no more values of n for which \[\left( -\pi \le x\le \pi \right)\].
(ii) Considering $\cos \left( \dfrac{x}{2} \right)=0$
Using the formula for general solution: $\cos x=0\Rightarrow x=\left( 2n+1 \right)\dfrac{\pi }{2}$, we get,
$\begin{align}
  & \dfrac{x}{2}=\left( 2n+1 \right)\dfrac{\pi }{2} \\
 & \Rightarrow x=\left( 2n+1 \right)\pi \\
\end{align}$
Now, we must take such integer values of n for which \[\left( -\pi \le x\le \pi \right)\]. Therefore,
For n = 0, we have,
$x=\pi$
For n = -1, we have,
$\begin{align}
  & x=\left( 2\times \left( -1 \right)+1 \right)\pi \\
 & \Rightarrow x=\left( -2+1 \right)\pi \\
 & \Rightarrow x=-\pi \\
\end{align}$
Now, there are no more values of n for which \[\left( -\pi \le x\le \pi \right)\].
(iii) Considering $\left( 1+\sin x \right)=0$
$\Rightarrow \sin x=-1$
The only value for which $\sin x=-1$ in the domain \[\left( -\pi \le x\le \pi \right)\] is $x=\dfrac{-\pi }{2}$.
Now, we can see that the solution $x=\pi$ is common in the 1st and 2nd case. Therefore, we will count it as one solution. Therefore, total solutions are: $-\pi ,\dfrac{-\pi }{3},\dfrac{-\pi }{2},\dfrac{\pi }{3}\text{ and }\pi$, which are 5 in numbers.
Hence, option (c) is the correct answer.

Note: One may note that we are not finding a general solution but the solutions in the domain: \[\left( -\pi \le x\le \pi \right)\]. So, we have to consider each case and find their solutions. In the above solution, there were three cases and we have found the values of x for those three cases one by one. Remember that if you are getting a similar value of x in two or more cases, count them together as one solution.