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Question: Solve \(cos\theta +\cos 2\theta +\cos 3\theta =0\)....

Solve cosθ+cos⁡2θ+cos⁡3θ=0cos\theta +\cos 2\theta +\cos 3\theta =0.

Explanation

Solution

Hint: Apply the identity of cosC⁡+cos⁡D\operatorname{cosC}+\cos D with the suitable terms of the given equation, which is given as cos⁡C+cos⁡D=2cos⁡C+D2cos⁡C−D2\cos C+\cos D=2\cos \dfrac{C+D}{2}\cos \dfrac{C-D}{2}
Now, use the general solution of the equationcos⁡x=coy\cos x=coy, after simplifying the given expression; which is given as:
If cos⁡x=cos⁡y\cos x=\cos y
Then general solution is given as
x=2nπ±yx=2n\pi \pm y Where n∈zn\in z

Complete step-by-step answer:
We have
cos⁡θ+cos⁡2θ+cos⁡3θ=0\cos \theta +\cos 2\theta +\cos 3\theta =0 ………….. (i)
As we know the trigonometric identity of cos⁡C+cos⁡D\cos C+\cos D can be given as
cosC⁡+cosD=2cosC+D2cos⁡C−D2\operatorname{cosC}+cosD=2cos\dfrac{C+D}{2}\cos \dfrac{C-D}{2} …………… (ii)
Now, we can observe the identity (ii) and get that if C + D and C - D will be even numbers then angles in the cosine functions will be of integer type otherwise, we get angles of them in fractions.
So, now we can observe equation (i) and get that we should add cos⁡θ,cos⁡3θ\cos \theta ,\cos 3\theta with the help of property (ii) as sum of θ,3θ\theta ,3\theta will be an even number. So, we can write equation (i) as:
(cos⁡θ+cos⁡3θ)+cos⁡2θ=0\left( \cos \theta +\cos 3\theta \right)+\cos 2\theta =0
Apply the identity in equation (ii) with the first two terms of the above equation. So, we get
2cos⁡(θ+3θ2)cos⁡(θ−3θ2)+cos⁡2θ=02\cos \left( \dfrac{\theta +3\theta }{2} \right)\cos \left( \dfrac{\theta -3\theta }{2} \right)+\cos 2\theta =0
2cos⁡(2θ)cos⁡(−θ)+cos⁡2θ=02\cos \left( 2\theta \right)\cos \left( -\theta \right)+\cos 2\theta =0
We know thatcos⁡(−θ)=cos⁡θ\cos \left( -\theta \right)=\cos \theta . So, we get
2cos⁡θcos⁡θ+cos⁡2θ=02\cos \theta \cos \theta +\cos 2\theta =0
Now, take ′cos⁡2θ′'\cos 2\theta ' as common from both the terms of the above equation and hence, we get
cos⁡2θ(2cos⁡θ+1)=0\cos 2\theta \left( 2\cos \theta +1 \right)=0 ………………(iii)
As we know that multiplication of two terms will be zero, if one of them will be 0 or both. So, we get from equation (iii) as
cos⁡2θ=0,2cos⁡θ+1=0\cos 2\theta =0,2\cos \theta +1=0
cos⁡2θ=0,cos⁡θ=−12\cos 2\theta =0,\cos \theta =\dfrac{-1}{2}…………… (iv)
We know that general solution of equation cos⁡x=cos⁡y\cos x=\cos y is given by relation
x=2nπ±yx=2n\pi \pm y ………… (v)
Where n∈zn\in z(integer)
So, we have
cos⁡2θ=0\cos 2\theta =0
As we know cos⁡θ\cos \theta gives value ‘0’ at θ=π2\theta =\dfrac{\pi }{2} , hence we get
cos⁡2θ=cos⁡π2\cos 2\theta =\cos \dfrac{\pi }{2}
Now, using relation (v), we get θ\theta as
2θ=2nπ±π22\theta =2n\pi \pm \dfrac{\pi }{2}
Divide the whole equation by 2, we get
2θ2=2nπ2±π4\dfrac{2\theta }{2}=\dfrac{2n\pi }{2}\pm \dfrac{\pi }{4}
θ=nπ±π4\theta =n\pi \pm \dfrac{\pi }{4}………………… (vi)
And we have another relation as
cos⁡θ=−12\cos \theta =\dfrac{-1}{2}
As we know that cos⁡θ\cos \theta will give 12\dfrac{1}{2} at θ=π3\theta =\dfrac{\pi }{3}and
We know the relationcos⁡(180−θ)=−cos⁡θ\cos \left( 180-\theta \right)=-\cos \theta . So, we can put θ=π3\theta =\dfrac{\pi }{3} here as well and hence, we get
cos⁡(180−θ)=cos⁡(π−θ)=−cos⁡θ cos⁡(π−π3)=−cos⁡π3=−12 \begin{aligned} & \cos \left( 180-\theta \right)=\cos \left( \pi -\theta \right)=-\cos \theta \\\ & \cos \left( \pi -\dfrac{\pi }{3} \right)=-\cos \dfrac{\pi }{3}=\dfrac{-1}{2} \\\ \end{aligned}
Or
cos⁡2π3=−12\cos \dfrac{2\pi }{3}=\dfrac{-1}{2}
It means cos⁡θ\cos \theta will give ‘−12\dfrac{-1}{2}’ at θ2π3\theta \dfrac{2\pi }{3}
Hence, we can write the relation cos⁡θ=−12\cos \theta =\dfrac{-1}{2} as cos⁡θ=cos⁡2π3\cos \theta =\cos \dfrac{2\pi }{3}
Hence, from the relation (v), we get
θ=2nπ±2π3\theta =2n\pi \pm \dfrac{2\pi }{3} …………….. (vii)
Hence, we get the solution of equation given in the problem (cos⁡θ+cos⁡2θ+cos⁡3θ=0)\left( \cos \theta +\cos 2\theta +\cos 3\theta =0 \right) as
θ=nπ±π4,θ=2nπ±2π3\theta =n\pi \pm \dfrac{\pi }{4},\theta =2n\pi \pm \dfrac{2\pi }{3}

Note: One may go wrong if he/she uses the given identity i.e. cos⁡C+cos⁡D=2cos⁡(C+D2)cos⁡(C−D2)\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right) with first two terms or the last two terms i.e. cos⁡θ+cos⁡2θ,cos⁡2θ+cos⁡3θ\cos \theta +\cos 2\theta ,\cos 2\theta +\cos 3\theta . As we will get a fraction from θ\theta inside the cosine functions and which will make the given relation complex. It means observing the given relations id the key point of the question.
One may go wrong if he/she solve the question in the following approach:
cos⁡θ+cos⁡3θ=−cos⁡2θ 2cos⁡2θcos⁡θ=−cos⁡2θ \begin{aligned} & \cos \theta +\cos 3\theta =-\cos 2\theta \\\ & 2\cos 2\theta \cos \theta =-\cos 2\theta \\\ \end{aligned}
Cancelling the terms cos⁡2θ\cos 2\theta from both sides, we get 2cos⁡2θ=−12\cos 2\theta =-1
So, he/she will miss the solution of the equationscos⁡2θ=0\cos 2\theta =0. So, take care of it and don’t cancel out the terms cos⁡2θ\cos 2\theta from both sides as the number of solutions will become less.