Question

How do you solve: \[2{{\cos }^{2}}\left( 4x \right)-1=0\]?

Answer 1

Hint: Assume the argument of the cosine function, i.e. 4x, equal to \[\theta \]. Now, use the half angle formula of the cosine function given as: - \[2{{\cos }^{2}}\theta -1=\cos 2\theta \] and simplify the given equation. Use the formula for general solution of a cosine function having the equation \[\cos \theta =0\] given as: - \[\theta =\left( 2n+1 \right)\dfrac{\pi }{2}\], where ‘n’ can be any integer.

Complete step by step solution:
Here, we have been provided with the trigonometric equation containing the cosine function given as: - \[2{{\cos }^{2}}\left( 4x \right)-1=0\] and we are asked to solve it. That means we need to find the values of x.
\[\because 2{{\cos }^{2}}\left( 4x \right)-1=0\]
Now, let us assume the argument of the cosine function, i.e., (4x), equal to \[\theta \], so we get,
\[\Rightarrow 2{{\cos }^{2}}\theta -1=0\]
Using the half angle formula of the cosine function given as: - \[2{{\cos }^{2}}\theta -1=\cos 2\theta \], we get,
\[\Rightarrow \cos 2\theta =0\] - (1)
We know that the value of the cosine function is 0 when we have an odd multiple of \[\dfrac{\pi }{2}\] as the angle in the argument of cosine function. Mathematically, we have the general solution given as: - if \[\cos 2\theta =0\] then \[\theta =\left( 2n+1 \right)\dfrac{\pi }{2}\], where \[n\in \] integers. So, for equation (1), we have,
\[\Rightarrow 2\theta =\left( 2n+1 \right)\dfrac{\pi }{2},n\in \] integers
Dividing both the sides with 2, we get,
\[\Rightarrow \theta =\left( 2n+1 \right)\dfrac{\pi }{4}\]
Substituting back the value of \[\theta =4x\], we get,
\[\Rightarrow 4x=\left( 2n+1 \right)\dfrac{\pi }{4}\]
Dividing both the sides with 4 to solve for the value of x, we get,
\[\Rightarrow x=\left( 2n+1 \right)\dfrac{\pi }{16},n\in \] integers
Hence, the solution of the given trigonometric equation is given as: - \[x=\left( 2n+1 \right)\dfrac{\pi }{16}\] where \[n\in \] integers.

Note: One may note that here we have found the general solution of the given equation. This is because we are not provided with any information regarding the interval of x between which we have to find the values. If any interval would have been provided then we would have substituted the suitable values of n to get the principal solutions. You can also solve the equation in a different manner by using the formula: - if \[{{\cos }^{2}}a={{\cos }^{2}}b\] then \[a=n\pi \pm b\]. In this case the solution may look different from what we have obtained but if you will substitute the values of n then you will get the same result.