Question

What is the solution of the equation $5\cos 2x + 1 = 3\cos 2x$ ?

Answer 1

Hint: We will simplify the equation by taking the trigonometric term to one side and the constant term by the other side. We will use the formula $\cos 2x = 2{\cos ^2}x - 1$ to further degrade the equation in terms of x. We should know the values of trigonometric ratios at some general points.

Complete step by step answer:
We have given the equation $5\cos 2x + 1 = 3\cos 2x$
$ \Rightarrow 5\cos 2x + 1 = 3\cos 2x$
We subtract 3cos2x on both side
$ \Rightarrow 5\cos 2x - 3\cos 2x = - 1$
$ \Rightarrow 2\cos 2x = - 1$
We have divided both side by 2
$ \Rightarrow \cos 2x = \dfrac{{ - 1}}{2}$
We know that $\cos 2x = 2{\cos ^2}x - 1$
$ \Rightarrow 2{\cos ^2}x - 1 = \dfrac{{ - 1}}{2}$
We add 1 on both side
$ \Rightarrow 2{\cos ^2}x = \dfrac{{ - 1}}{2} + 1$
$ \Rightarrow 2{\cos ^2}x = \dfrac{1}{2}$
We have divided both side by 2
$ \Rightarrow {\cos ^2}x = \dfrac{1}{4}$
$ \Rightarrow \cos x = \pm \dfrac{1}{2}$
We know that cos x repeats itself in interval of $2\pi $
 So, the value of x is $x = \dfrac{\pi }{3} + 2n\pi $ and $x = \dfrac{{5\pi }}{3} + 2n\pi $ for all integral values of n.
Hence, the solution of $5\cos 2x + 1 = 3\cos 2x$ is $x = \dfrac{\pi }{3} + 2n\pi $ and $x = \dfrac{{5\pi }}{3} + 2n\pi $

Note:We know that sin x and cos x repeat after an $2\pi $ interval, whereas tan x repeats after a $\pi $ interval which is also called their periodicity. Principal solutions are solutions to trigonometry equations that fall inside the range 0 and $2\pi $.