
Find the general solution of cos4x = cos2x
Answer
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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 $\cos C-\cos D=-2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)$ , 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.
$\begin{align}
& \cos 4x=\cos 2x \\
& \cos 4x-\cos 2x=0 \\
\end{align}$
Now using the formula $\cos C-\cos D=-2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)$ we get,
$-2\sin \left( \dfrac{4x+2x}{2} \right)\sin \left( \dfrac{4x-2x}{2} \right)=0$
$-2\sin 3x\sin x=0$
From this we get two equations,
$\sin 3x=0\text{ or }\sin x=0$
Let’s first solve for sin3x = 0,
We know that sin0 = 0,
Therefore we get,
sin3x = sin0
Now, if we have $\sin \theta =\sin \alpha $ then the general solution is:
$\theta =n\pi +{{\left( -1 \right)}^{n}}\alpha $
Now using the above formula for sin3x = sin0 we get,
$\begin{align}
& 3x=n\pi +{{\left( -1 \right)}^{n}}0 \\
& x=\dfrac{n\pi }{3}..........(1) \\
\end{align}$
Now we will solve sinx = 0,
We know that sin0 = 0,
Therefore we get,
sinx = sin0
Now, if we have $\sin \theta =\sin \alpha $ then the general solution is:
$\theta =n\pi +{{\left( -1 \right)}^{n}}\alpha $
Now using the above formula for sin3x = sin0 we get,
$\begin{align}
& x=n\pi +{{\left( -1 \right)}^{n}}0 \\
& x=n\pi ..........(2) \\
\end{align}$
Now from equation (1) and (2) we can say that the answer is,
$x=\dfrac{n\pi }{3}\text{ or }x=n\pi $
Hence, this is the answer to this question.
Note: The trigonometric formula $\cos C-\cos D=-2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)$ and the formula for general solution of sin must be kept in mind. In this question we can also take the value of $\alpha $ as $\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.
Complete step-by-step answer:
Let’s start solving the question.
$\begin{align}
& \cos 4x=\cos 2x \\
& \cos 4x-\cos 2x=0 \\
\end{align}$
Now using the formula $\cos C-\cos D=-2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)$ we get,
$-2\sin \left( \dfrac{4x+2x}{2} \right)\sin \left( \dfrac{4x-2x}{2} \right)=0$
$-2\sin 3x\sin x=0$
From this we get two equations,
$\sin 3x=0\text{ or }\sin x=0$
Let’s first solve for sin3x = 0,
We know that sin0 = 0,
Therefore we get,
sin3x = sin0
Now, if we have $\sin \theta =\sin \alpha $ then the general solution is:
$\theta =n\pi +{{\left( -1 \right)}^{n}}\alpha $
Now using the above formula for sin3x = sin0 we get,
$\begin{align}
& 3x=n\pi +{{\left( -1 \right)}^{n}}0 \\
& x=\dfrac{n\pi }{3}..........(1) \\
\end{align}$
Now we will solve sinx = 0,
We know that sin0 = 0,
Therefore we get,
sinx = sin0
Now, if we have $\sin \theta =\sin \alpha $ then the general solution is:
$\theta =n\pi +{{\left( -1 \right)}^{n}}\alpha $
Now using the above formula for sin3x = sin0 we get,
$\begin{align}
& x=n\pi +{{\left( -1 \right)}^{n}}0 \\
& x=n\pi ..........(2) \\
\end{align}$
Now from equation (1) and (2) we can say that the answer is,
$x=\dfrac{n\pi }{3}\text{ or }x=n\pi $
Hence, this is the answer to this question.
Note: The trigonometric formula $\cos C-\cos D=-2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)$ and the formula for general solution of sin must be kept in mind. In this question we can also take the value of $\alpha $ as $\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.
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