Question

Solve the equation \[\cos \theta +cos3\theta -\cos 2\theta =0\].

Answer 1

Hint: We will apply the trigonometric formula given by \[\cos 2\theta =2{{\cos }^{2}}\theta -1\] and \[\cos 3\theta =4co{{s}^{3}}\theta -3\cos \theta \] where \[2\theta \] and \[3\theta \] are twice and thrice of the angle \[\theta \]. After applying these, the given equation will be simplified as an equation of degree 3, which can then be solved further.

Complete step-by-step answer:

Consider equation, \[\cos \theta +cos3\theta -\cos 2\theta =0.....(1)\]
Now we will apply the formula of \[\cos 2\theta =2{{\cos }^{2}}\theta -1\] and \[\cos 3\theta =4co{{s}^{3}}\theta -3\cos \theta \] in equation (1).
\[\begin{align}
  & \Rightarrow \cos \theta +\left( 4co{{s}^{3}}\theta -3\cos \theta \right)-\left( 2{{\cos }^{2}}\theta -1 \right)=0 \\
 & \Rightarrow \cos \theta +4{{\cos }^{3}}\theta -3\cos \theta -2{{\cos }^{2}}\theta +1=0 \\
\end{align}\]
Now we will arrange the expression in the order of degree of \[cos\theta \]. That is,
\[\begin{align}
  & \Rightarrow 4{{\cos }^{3}}\theta -2{{\cos }^{2}}\theta -3\cos \theta +\cos \theta +1=0 \\
 & \Rightarrow 4{{\cos }^{3}}\theta -2{{\cos }^{2}}\theta -2\cos \theta +1=0...(2) \\
\end{align}\]
Now we will put \[x=\cos \theta \] in equation (2). This will result into \[{{x}^{2}}={{\cos }^{2}}\theta \] and \[{{x}^{3}}={{\cos }^{3}}\theta \]. So we get a new expression, given by,
\[\begin{align}
  & 4{{\cos }^{3}}\theta -2{{\cos }^{2}}\theta +1=0 \\
 & 4{{x}^{3}}-2{{x}^{2}}-2x+1=0....(3) \\
\end{align}\]
Now we will use the hit and trial method and substitute the value of x in equation (3). First substitute x = 0 in equation (3).
\[4{{\left( 0 \right)}^{3}}-2{{\left( 0 \right)}^{2}}-2\left( 0 \right)+1=1\]
This value of x = 0 does not satisfy the equation (3) as the equation does not result into 0. Now we will apply the value \[x=\dfrac{1}{2}\] and substitute it in equation (3).
\[\Rightarrow 4{{\left( \dfrac{1}{2} \right)}^{3}}-2{{\left( \dfrac{1}{2} \right)}^{2}}-2\left( \dfrac{1}{2} \right)+1=\dfrac{4}{8}-\dfrac{2}{4}-1+1=0\]
This clearly implies that \[x=\dfrac{1}{2}\] is a factor of equation (3) or \[\left( x-\dfrac{1}{2} \right)\] is a factor.
Now we will divide \[x-\dfrac{1}{2}\] to equation (3).
Thus we get,
\[x-\dfrac{1}{2}\overset{4{{x}^{2}}-2}{\overline{\left){\begin{align}
  & 4{{x}^{3}}-2{{x}^{2}}-2x+1 \\
 & \underline{4{{x}^{3}}-2{{x}^{2}}} \\
 & -2x+1 \\
 & \underline{-2x+1} \\
 & \underline{000000} \\
\end{align}}\right.}}\]
Clearly we factorize the equation \[4{{x}^{3}}-2{{x}^{2}}-2x+1\] into \[\left( x-\dfrac{1}{2} \right)\left( 4{{x}^{2}}-2 \right)\].
\[\left( x-\dfrac{1}{2} \right)\left( 4{{x}^{2}}-2 \right)=0\]
Now \[x-\dfrac{1}{2}=0\] or \[4{{x}^{2}}-2=0\].
Consider \[x-\dfrac{1}{2}=0\] or \[x=\dfrac{1}{2}\] since we know \[x=\cos \theta \]. \[\Rightarrow \cos \theta =\dfrac{1}{2}\].
We know that the \[\dfrac{1}{2}\] is the value of \[\cos \dfrac{\pi }{3}\].
\[\Rightarrow \cos \theta =\dfrac{\pi }{3}\].
The value of cos is positive in the first and fourth quadrant. In quadrant I, we have \[\cos \theta =\dfrac{\pi }{3}\].
\[\begin{align}
  & \Rightarrow \cos \theta =\dfrac{\pi }{3} \\
 & \Rightarrow \theta =\dfrac{\pi }{3} \\
\end{align}\]
\[\Rightarrow \theta =\dfrac{\pi }{3}+2n\pi \], where \[n=0,\pm 1,\pm 2....\]
In fourth quadrant, we have,
\[\begin{align}
  & \cos \dfrac{\pi }{3}=\cos \left( 2\pi -\dfrac{\pi }{3} \right) \\
 & \cos \theta =\cos \left( \dfrac{6\pi -\pi }{3} \right) \\
 & \cos \theta =\cos \left( \dfrac{5\pi }{3} \right) \\
 & \theta =\dfrac{5\pi }{3} \\
\end{align}\]
\[\theta =\dfrac{5\pi }{3}+2n\pi \], where \[n=0,\pm 1,\pm 2....\]
Now we will consider \[4{{x}^{2}}-2=0\].
\[\begin{align}
  & 4{{x}^{2}}=2 \\
 & {{x}^{2}}=\dfrac{1}{2} \\
 & x=\pm \dfrac{1}{\sqrt{2}} \\
\end{align}\]
Since we know that \[\cos \theta =x\]. Thus we can write \[\cos \theta =\dfrac{1}{\sqrt{2}}\] and \[\cos \theta =\dfrac{-1}{\sqrt{2}}\].
Now we will consider, \[\cos \theta =\dfrac{1}{\sqrt{2}}\]. As we know that \[\cos \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}\].
\[\Rightarrow \cos \theta =\cos \dfrac{\pi }{4}\]
The value of cos is positive in the first and fourth quadrant. So we first consider quadrant I.
\[\begin{align}
  & \Rightarrow \cos \theta =\cos \dfrac{\pi }{4} \\
 & \Rightarrow \theta =\dfrac{\pi }{4} \\
\end{align}\]
\[\Rightarrow \theta =\dfrac{\pi }{4}+2n\pi \], where \[n=0,\pm 1,\pm 2....\]
Now we will consider the fourth quadrant.
In the fourth quadrant, the value of \[\cos \dfrac{\pi }{4}=\cos \left( 2\pi -\dfrac{\pi }{4} \right)\].
\[\begin{align}
  & \Rightarrow \cos \theta =\cos \left( 2\pi -\dfrac{\pi }{4} \right) \\
 & \Rightarrow \cos \theta =\cos \left( \dfrac{8\pi -\pi }{4} \right) \\
 & \Rightarrow \cos \theta =\cos \left( \dfrac{7\pi }{4} \right) \\
 & \Rightarrow \theta =\dfrac{7\pi }{4} \\
\end{align}\]
\[\Rightarrow \theta =\dfrac{7\pi }{4}+2n\pi \], where \[n=0,\pm 1,\pm 2....\]
Now we will consider \[\cos \theta =\dfrac{-1}{\sqrt{2}}\] as the value of \[\dfrac{1}{\sqrt{2}}=\cos \dfrac{\pi }{4}\].
\[\Rightarrow \cos \theta =-\cos \dfrac{\pi }{4}\]
We know that the value of cos is negative in the second and third quadrant. So we will consider the second quadrant. In this quadrant, the value of \[-\cos \dfrac{\pi }{4}=\cos \left( \pi -\dfrac{\pi }{4} \right)\].
\[\begin{align}
  & \Rightarrow cos\theta =cos\left( \pi -\dfrac{\pi }{4} \right) \\
 & \Rightarrow cos\theta =cos\left( \dfrac{3\pi }{4} \right) \\
 & \Rightarrow \theta =\dfrac{3\pi }{4} \\
\end{align}\]
\[\Rightarrow \theta =\dfrac{3\pi }{4}+2n\pi \], where \[n=0,\pm 1,\pm 2....\]
Also, in the third quadrant the value of \[-\cos \dfrac{\pi }{4}=\cos \left( \pi +\dfrac{\pi }{4} \right)\].
\[\begin{align}
  & \Rightarrow \cos \theta =\cos \left( \pi +\dfrac{\pi }{4} \right) \\
 & \Rightarrow \cos \theta =\cos \left( \dfrac{5\pi }{4} \right) \\
 & \Rightarrow \theta =\dfrac{5\pi }{4} \\
\end{align}\]
\[\Rightarrow \theta =\dfrac{5\pi }{4}+2n\pi \], where \[n=0,\pm 1,\pm 2....\]
Hence the solution of equation (1) is given by,
\[\theta =\dfrac{\pi }{3},\dfrac{5\pi }{3},\dfrac{\pi }{4},\dfrac{7\pi }{4},\dfrac{3\pi }{4},\dfrac{5\pi }{4}.\]

Note: Alternatively, we could have factorized the equation \[4{{x}^{2}}-2\] by factorization Actually, we could have used factorization method in the equation \[4{{x}^{3}}-2{{x}^{2}}-2x+1\]. Since we are asked about the basic solution instead of the general solutions that why we write \[\theta =\dfrac{\pi }{2},\theta =\dfrac{2\pi }{3}\] and so on instead of \[\theta =\dfrac{\pi }{2}+2n\pi ,\theta =\dfrac{2\pi }{3}+2n\pi \].