Question

Solve using the double angle understanding? $\cos 6x + \cos 4x + \cos 2x = 0$

Answer 1

Hint: here we are asked to solve the given equation which has trigonometric functions. Also, we are asked to use the double angle concept. As we can see that the given equation contains only cosine function we will use the double formula of cosine functions that is $\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$ and $\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$ by modulating the angle of these formulae some formulae are derived those formulae are used in the problem to solve the given equation.
Formula to be used:
a) $\cos a + \cos b = 2\cos \left( {\dfrac{{a + b}}{2}} \right)\cos \left( {\dfrac{{a - b}}{2}} \right)$
b) When $\cos x = \cos \theta $ then the general solution is given by $x = 2n\pi \pm \theta $ , where $\theta \in \left( {0,\pi } \right]$

Complete answer:
The given equation is $\cos 6x + \cos 4x + \cos 2x = 0$and we need to solve it (i.e. we need to calculate the solutions for the given trigonometric equation).
Let us number the given equation $\cos 6x + \cos 4x + \cos 2x = 0$as $\left( 1 \right)$
Let us consider $\cos 6x + \cos 2x$
Now, we shall apply the formula $\cos a + \cos b = 2\cos \left( {\dfrac{{a + b}}{2}} \right)\cos \left( {\dfrac{{a - b}}{2}} \right)$
Thus, we have $\cos 6x + \cos 2x$$ = 2\cos \left( {\dfrac{{6x + 2x}}{2}} \right)\cos \left( {\dfrac{{6x - 2x}}{2}} \right)$
$ = 2\cos \left( {\dfrac{{8x}}{2}} \right)\cos \left( {\dfrac{{4x}}{2}} \right)$
$ = 2\cos 4x\cos 2x$
Hence, we get $\cos 6x + \cos 2x$$ = 2\cos 4x\cos 2x$……….$\left( 2 \right)$
Now, we shall substitute the equation $\left( 2 \right)$in the equation $\left( 1 \right)$
Thus, we will get $\cos 6x + \cos 4x + \cos 2x = 0$
$ \Rightarrow \cos 4x + 2\cos 4x\cos 2x = 0$
$ \Rightarrow \cos 4x\left( {1 + 2\cos 2x} \right) = 0$
Hence in the above equation, we can note that either $\cos 4x = 0$ or $1 + 2\cos 2x = 0$ .
We shall consider both cases to find the solution.
Case a:
Let us consider $\cos 4x = 0$
$ \Rightarrow \cos 4x = \cos 90^\circ $ (We know that $\cos 90^\circ = 0$ )
$ \Rightarrow \cos 4x = \cos \dfrac{\pi }{2}$
$ \Rightarrow 4x = 2n\pi \pm \dfrac{\pi }{2}$ (Here we have applied $\cos x = \cos \theta $$ \Rightarrow x = 2n\pi \pm \theta $ , where $\theta \in \left( {0,\pi } \right]$)
$ \Rightarrow x = \dfrac{{2n\pi }}{4} \pm \dfrac{\pi }{{2 \times 4}}$
 $ \Rightarrow x = \dfrac{{n\pi }}{2} \pm \dfrac{\pi }{8}$
Thus, we got the solution $x = \dfrac{{n\pi }}{2} \pm \dfrac{\pi }{8}$ when $\cos 4x = 0$for the given equation.
Case b:
Let us consider $1 + 2\cos 2x = 0$
$ \Rightarrow 2\cos 2x = - 1$
$ \Rightarrow \cos 2x = - \dfrac{1}{2}$
$ \Rightarrow \cos 2x = \cos 120^\circ $ (We know that $\cos 120^\circ = - \dfrac{1}{2}$ )
$ \Rightarrow \cos 2x = \cos \dfrac{{2\pi }}{3}$
$ \Rightarrow 2x = 2n\pi \pm \dfrac{{2\pi }}{3}$ (Here we have applied $\cos x = \cos \theta $$ \Rightarrow x = 2n\pi \pm \theta $ , where $\theta \in \left( {0,\pi } \right]$)
$ \Rightarrow x = \dfrac{{2n\pi }}{2} \pm \dfrac{{2\pi }}{{3 \times 2}}$
 $ \Rightarrow x = n\pi \pm \dfrac{\pi }{3}$
Thus, we got the solution $x = n\pi \pm \dfrac{\pi }{3}$ when $1 + 2\cos 2x = 0$for the given equation.

Note:
Since we are asked to solve the given equation, we have calculated the appropriate solution of the given equation. Also, it is no need to have a single solution for an equation. An equation can contain more than one solution and even no solution can also exist. Here, we can have either $x = \dfrac{{n\pi }}{2} \pm \dfrac{\pi }{8}$or $x = n\pi \pm \dfrac{\pi }{3}$for the given equation.