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Question

Question: Find the general solution of cos4x = cos2x...

Find the general solution of cos4x = cos2x

Explanation

Solution

Hint: First we will rearrange the given equation of taking the variable x to one side and after that we will use the trigonometric formula cosCcosD=2sin(C+D2)sin(CD2)\cos C-\cos D=-2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right) , and then we will use the formula for finding the general solution of two different equation of sin, and that will be the answer.

Complete step-by-step answer:
Let’s start solving the question.
cos4x=cos2x cos4xcos2x=0 \begin{aligned} & \cos 4x=\cos 2x \\\ & \cos 4x-\cos 2x=0 \\\ \end{aligned}
Now using the formula cosCcosD=2sin(C+D2)sin(CD2)\cos C-\cos D=-2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right) we get,
2sin(4x+2x2)sin(4x2x2)=0-2\sin \left( \dfrac{4x+2x}{2} \right)\sin \left( \dfrac{4x-2x}{2} \right)=0
2sin3xsinx=0-2\sin 3x\sin x=0
From this we get two equations,
sin3x=0 or sinx=0\sin 3x=0\text{ or }\sin x=0
Let’s first solve for sin3x = 0,
We know that sin0 = 0,
Therefore we get,
sin3x = sin0
Now, if we have sinθ=sinα\sin \theta =\sin \alpha then the general solution is:
θ=nπ+(1)nα\theta =n\pi +{{\left( -1 \right)}^{n}}\alpha
Now using the above formula for sin3x = sin0 we get,
3x=nπ+(1)n0 x=nπ3..........(1) \begin{aligned} & 3x=n\pi +{{\left( -1 \right)}^{n}}0 \\\ & x=\dfrac{n\pi }{3}..........(1) \\\ \end{aligned}
Now we will solve sinx = 0,
We know that sin0 = 0,
Therefore we get,
sinx = sin0
Now, if we have sinθ=sinα\sin \theta =\sin \alpha then the general solution is:
θ=nπ+(1)nα\theta =n\pi +{{\left( -1 \right)}^{n}}\alpha
Now using the above formula for sin3x = sin0 we get,
x=nπ+(1)n0 x=nπ..........(2) \begin{aligned} & x=n\pi +{{\left( -1 \right)}^{n}}0 \\\ & x=n\pi ..........(2) \\\ \end{aligned}
Now from equation (1) and (2) we can say that the answer is,
x=nπ3 or x=nπx=\dfrac{n\pi }{3}\text{ or }x=n\pi
Hence, this is the answer to this question.

Note: The trigonometric formula cosCcosD=2sin(C+D2)sin(CD2)\cos C-\cos D=-2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right) and the formula for general solution of sin must be kept in mind. In this question we can also take the value of α\alpha as π\pi , and then use the formula of general solution to find the answer, the answer that we get is also correct, so if there are multiple options then one can make a mistake thinking that only one answer is correct.