Question

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

Answer 1

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.