Question

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

Answer 1

Hint: We will apply the formula of \[cos2\theta =2{{\cos }^{2}}\theta -1\] and \[\cos 3\theta =4{{\cos }^{3}}\theta -3\cos \theta \]. These are trigonometric formulas for twice and thrice of angle respectively. After using this, we will get an equation of degree 3, which can then be solved further.

Complete step-by-step answer:

We will first consider the equation \[\cos \theta +\cos 2\theta +cos3\theta =0......(1)\]
Now we will apply the trigonometric formula of \[cos2\theta =2{{\cos }^{2}}\theta -1\]and \[\cos 3\theta =4{{\cos }^{3}}\theta -3\cos \theta \] in equation (1). Therefore, we get a following new expression which is given by,
\[\begin{align}
  & \cos \theta +\left( 2{{\cos }^{2}}\theta -1 \right)+\left( 4{{\cos }^{3}}\theta -3\cos \theta \right)=0 \\
 & \Rightarrow \cos \theta +2{{\cos }^{2}}\theta -1+4{{\cos }^{3}}\theta -3\cos \theta =0 \\
\end{align}\]
Now we will arrange this equation in ascending order of the degree of \[\cos \theta \]. So now we have that,
\[\begin{align}
  & 4{{\cos }^{3}}\theta +2{{\cos }^{2}}\theta -3\cos \theta +\cos \theta -1=0 \\
 & 4{{\cos }^{3}}\theta +2{{\cos }^{2}}\theta -2\cos \theta -1=0 \\
\end{align}\]
Now we will substitute the trigonometric term \[\cos \theta =x\].
This will result into \[{{\cos }^{3}}\theta ={{x}^{3}},{{\cos }^{2}}\theta ={{x}^{2}},\cos \theta =x\]. Therefore, we get as follow
\[4{{x}^{3}}+2{{x}^{2}}-2x-1=0.....(2)\]
By hit and trial method, we will find the first factor of this equation (2) First substitute x = 0 and then check if the left hand side of the expression is equal to 0 or not.
\[4{{\left( 0 \right)}^{3}}+2{{\left( 0 \right)}^{2}}-2\left( 0 \right)-1=-1\] and this is not equal to 0. So, x = 0 is not a factor of equation (2). Now we will substitute \[x=\dfrac{1}{2}\] in equation (2).
\[\begin{align}
  & 4{{\left( \dfrac{1}{2} \right)}^{3}}+2{{\left( \dfrac{1}{2} \right)}^{2}}-2\left( \dfrac{1}{2} \right)-1=\dfrac{4}{8}+\dfrac{2}{4}-1-1 \\
 & \dfrac{4}{8}+\dfrac{2}{4}-1-1=\dfrac{1}{2}+\dfrac{1}{2}-1=\dfrac{1}{2}\times 2-2=-1 \\
\end{align}\]
This does not satisfy the equation (2). Now we will substitute \[x=\dfrac{-1}{2}\] in equation (2).
\[\begin{align}
  & 4{{\left( \dfrac{-1}{2} \right)}^{3}}+2{{\left( \dfrac{-1}{2} \right)}^{2}}-2\left( \dfrac{-1}{2} \right)-1=\dfrac{-4}{8}+\dfrac{2}{4}+\dfrac{2}{2}-1 \\
 & \dfrac{-4}{8}+\dfrac{2}{4}+\dfrac{2}{2}-1=\dfrac{-1}{2}+\dfrac{1}{2}+1-1 \\
 & \dfrac{-1}{2}+\dfrac{1}{2}+1-1=0 \\
\end{align}\]
This clearly satisfies equation (2). So we have \[x=\dfrac{-1}{2}\]as the first factor of equation (2). Now we will divide equation (2) by \[\left( x+\dfrac{1}{2} \right)\] which is its first factor.
Therefore, we have \[\left( x+\dfrac{1}{2} \right)\] as \[\left( 2x+1 \right)\] results into,
\[2x+1\overset{2{{x}^{2}}-1}{\overline{\left){\begin{align}
  & 4{{x}^{3}}-2{{x}^{2}}-2x+1 \\
 & \underline{4{{x}^{3}}-2{{x}^{2}}} \\
 & -2x-1 \\
 & \underline{-2x-1} \\
 & \underline{000000} \\
\end{align}}\right.}}\]
Clearly we have,
\[4{{x}^{3}}+2{{x}^{2}}-2x-1=\left( 2x+1 \right)\left( 2{{x}^{2}}-1 \right)\] which is equal to 0.
\[\begin{align}
  & \Rightarrow \left( 2x+1 \right)\left( 2{{x}^{2}}-1 \right)=0 \\
 & \Rightarrow 2x+1=0\Rightarrow 2{{x}^{2}}-1=0 \\
\end{align}\]
If \[2x+1=0\], therefore it results in \[x=\dfrac{-1}{2}\].
Since we know that \[x=\cos \theta \]. Thus we have \[\cos \theta =\dfrac{-1}{2}\]. By the value of \[\cos \dfrac{\pi }{3}=\dfrac{1}{2}\] we get \[\cos \theta =\dfrac{-1}{2}\]. By the value of \[\cos \dfrac{\pi }{3}=\dfrac{1}{2}\] we get \[\cos \theta =-\cos \dfrac{\pi }{3}\] by the fact that cos is always negative in second and third quadrant. Therefore, we have two expressions to which \[-\cos \dfrac{\pi }{3}\] will be equal.
We consider the second quadrant. Now we have \[-\cos \dfrac{\pi }{3}=\cos \left( \pi -\dfrac{\pi }{3} \right)\].
\[\begin{align}
  & \Rightarrow \cos \theta =\cos \left( \pi -\dfrac{\pi }{3} \right) \\
 & \Rightarrow \cos \theta =\cos \left( \dfrac{3\pi -\pi }{3} \right) \\
 & \Rightarrow \cos \theta =\cos \left( \dfrac{2\pi }{3} \right) \\
 & \Rightarrow \theta =\dfrac{2\pi }{3} \\
\end{align}\]
\[\Rightarrow \theta =\dfrac{2\pi }{3}+2n\pi \], where \[n=0,\pm 1,\pm 2....\]
Now we consider the third quadrant. We get \[-\cos \dfrac{\pi }{3}=\cos \left( \pi +\dfrac{\pi }{3} \right)\].
\[\begin{align}
  & \Rightarrow \cos \theta =\cos \left( \pi +\dfrac{\pi }{3} \right) \\
 & \Rightarrow \cos \theta =\cos \left( \dfrac{3\pi +\pi }{3} \right) \\
 & \Rightarrow \cos \theta =\cos \dfrac{4\pi }{3} \\
 & \Rightarrow \theta =\dfrac{4\pi }{3} \\
\end{align}\]
\[\Rightarrow \theta =\dfrac{4\pi }{3}+2n\pi \], where \[n=0,\pm 1,\pm 2....\]
Now if we consider \[2{{x}^{2}}-1=0\].
\[\begin{align}
  & {{x}^{2}}=\dfrac{1}{2} \\
 & x=\pm \dfrac{1}{\sqrt{2}} \\
\end{align}\]
Since we know that \[x=\cos \theta \]. Therefore, we have \[\cos \theta =\pm \dfrac{1}{\sqrt{2}}\] separately.
Consider \[\cos \theta =\pm \dfrac{1}{\sqrt{2}}\]. We know that,
\[\begin{align}
  & \dfrac{1}{\sqrt{2}}=\cos \dfrac{\pi }{4} \\
 & \Rightarrow \cos \theta =\cos \dfrac{\pi }{4} \\
\end{align}\]
Now we know that the value of cos is positive in the first and fourth quadrants. Therefore, we consider the first quadrant first and then the fourth quadrant.
In the first quadrant, we have \[\cos \theta =\cos \dfrac{\pi }{4}\].
\[\Rightarrow \theta =\dfrac{\pi }{4}\]
\[\Rightarrow \theta =\dfrac{\pi }{4}+2n\pi \], where \[n=0,\pm 1,\pm 2....\]
In the fourth quadrant, we have,
\[\begin{align}
  & \Rightarrow cos\dfrac{\pi }{4}=\cos \left( 2\pi -\dfrac{\pi }{4} \right) \\
 & \Rightarrow \cos \theta =\cos \left( \dfrac{8\pi -\pi }{4} \right) \\
 & \Rightarrow \cos \theta =\cos \left( \dfrac{7\pi }{4} \right) \\
 & \Rightarrow \theta =\dfrac{7\pi }{4} \\
\end{align}\]
\[\Rightarrow \theta =\dfrac{7\pi }{4}+2n\pi \], where \[n=0,\pm 1,\pm 2....\]
Now we will consider, \[\cos \theta =\dfrac{-1}{\sqrt{2}}\]. We know that,
\[\begin{align}
  & \dfrac{-1}{\sqrt{2}}=-\left( \cos \dfrac{\pi }{4} \right) \\
 & \cos \theta =-\left( \cos \dfrac{\pi }{4} \right) \\
\end{align}\]
Now we know that the value of cos is negative in the second and third quadrant. In this quadrant, we have,
\[\begin{align}
  & -\cos \dfrac{\pi }{4}=\cos \left( \pi -\dfrac{\pi }{4} \right) \\
 & \cos \theta =\cos \left( \pi -\dfrac{\pi }{4} \right) \\
 & \cos \theta =\cos \left( \dfrac{4\pi -\pi }{4} \right) \\
 & \cos \theta =\cos \dfrac{3\pi }{4} \\
 & \theta =\dfrac{3\pi }{4} \\
\end{align}\]
\[\theta =\dfrac{3\pi }{4}+2n\pi \], where \[n=0,\pm 1,\pm 2....\]
Also, in third quadrant, we have,
\[\begin{align}
  & -\cos \dfrac{\pi }{4}=\cos \left( \pi +\dfrac{\pi }{4} \right) \\
 & \cos \theta =\cos \left( \pi +\dfrac{\pi }{4} \right) \\
 & \cos \theta =\cos \left( \dfrac{4\pi +\pi }{4} \right) \\
 & \cos \theta =\cos \dfrac{5\pi }{4} \\
 & \theta =\dfrac{5\pi }{4} \\
\end{align}\]
\[\theta =\dfrac{5\pi }{4}+2n\pi \], where \[n=0,\pm 1,\pm 2....\]
Hence, the solution of equation (2) is given by,
\[\theta =\dfrac{2\pi }{3},\dfrac{4\pi }{3},\dfrac{\pi }{4},\dfrac{7\pi }{4},\dfrac{3\pi }{4},\dfrac{5\pi }{4}.\]

Note: We could have used factorization in the equation \[\cos \theta +\cos 2\theta +cos3\theta =0\], after solving equal to \[4{{x}^{3}}+2{{x}^{2}}-2x-1=0\]. But it will take time and make calculations complex. Since, we are asked about the basic solutions instead of general solutions that is why we write \[\theta =\dfrac{\pi }{2},\theta =\dfrac{2\pi }{3}\] instead of \[\theta =\dfrac{\pi }{2}+2n\pi ,\theta =\dfrac{2\pi }{3}+2n\pi \] and so on.