Question

Prove that \[\cos 3\theta = \cos 2\theta \] (Find general solution)

Answer 1

Hint: We have to find the general solution of\[\cos 3\theta = \cos 2\theta \]. For this function on the left hand side. When we apply formulas to the function. This will give us a particular value of the function by equation to zero. After that we can find the general solution.

Complete step-by-step answer:
We have given that \[\cos 3\theta = \cos 2\theta \]
Taking \[\cos 2\theta = 0\]
Now we have a trigonometric identity \[\cos A - \cos B\]=\[ - \sin \dfrac{{A + B}}{2}\]
Replacing \[A\]from \[3\theta \] and \[B\]from \[2\theta \]in the identity we get
\[\cos 3\theta - \cos 2\theta = - 2\sin \dfrac{{3\theta + 2\theta }}{2}.\sin \dfrac{{3\theta - 2\theta }}{2}\]
So, \[ - 2\sin \dfrac{{5\theta }}{2}.\sin \dfrac{\theta }{2} = 0\]
$\Rightarrow$ \[\sin \dfrac{{5\theta }}{2}.\sin \dfrac{\theta }{2} = \dfrac{0}{2} = 0\]
$\Rightarrow$ \[\sin \dfrac{{5\theta }}{2}.\sin \dfrac{\theta }{2} = 0\]
Now product of two functions is equal to 0
Therefore either \[\dfrac{{\sin 5\theta }}{2} = 0\] or \[\dfrac{{\sin \theta }}{2} = 0\] separately
Solving for \[\dfrac{{\sin 5\theta }}{2} = 0\] :
We have \[\dfrac{{\sin 5\theta }}{2} = 0\]
$\Rightarrow$ \[\dfrac{{5\theta }}{2} = \] $0$, $\pi $ , $2\pi $, $3\pi $...... when traced out in positive direction -----(i)
and \[\dfrac{{5\theta }}{2} = \] $ - \pi $ , $ - 2\pi $, $ - 3\pi $...... when traced out in negative direction -----(ii)
Combining (i) ad (ii)
$\Rightarrow$ \[\dfrac{{5\theta }}{2} = \]…………… $ - 3\pi $, $ - 2\pi $, $ - \pi $, $0$, $\pi $ , $2\pi $, $3\pi $………
$\Rightarrow$ \[\dfrac{{5\theta }}{2} = \]…………… $n\pi $, when n is integer
$\Rightarrow$ \[\theta = \dfrac{{2n\pi }}{5}\]
Solving for \[\dfrac{{\sin \theta }}{2} = 0\] :
We have \[\dfrac{{\sin \theta }}{2} = 0\]
$\Rightarrow$ \[\dfrac{\theta }{2} = \] $n\pi $ when n is an integer\[\theta = 2n\pi \]
Therefore general solution for \[\cos 3\theta = \cos 2\theta \] as \[\dfrac{{2n\pi }}{5},2n\pi \] When n is an integer .


Note: General Solution: - The equation that involves the trigonometric function of a variable is called Trigonometric equation. These equations have one or more trigonometric ratios of unknown angles.
For example \[\cos x - {\sin ^2}x = 0,\]is a trigonometric equation which does not satisfy all the values of x. Hence for such an equation we have to find the value of\[x\].
We know that \[\sin x\]and \[\cos x\]repeats themselves after an interval \[2\pi \] and \[\tan x\] repeats itself after interval\[\pi \]. The solution of such trigonometry function which is in interval \[[0,2\pi ]\] is called principal solution. A trigonometry equation will also have a general solution satisfying the equation.