Question

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

Answer 1

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 solution:
We have given that $\cos 3\theta = \cos 2\theta$
Taking $\cos 2\theta$ to the left hand side.
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, $- 2\sin \dfrac{{5\theta }}{2}.\sin \dfrac{\theta }{2} = 0$
$\sin \dfrac{{5\theta }}{2}.\sin \dfrac{\theta }{2} = \dfrac{0}{2} = 0$
$\sin \dfrac{{5\theta }}{2}.\sin \dfrac{\theta }{2} = 0$
Now the 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 $\dfrac{{\sin 5\theta }}{2} = 0$
$\dfrac{{5\theta }}{2} =$ $0$, $\pi$ , $2\pi$, $3\pi$...... 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)
$\dfrac{{5\theta }}{2} =$…………… $- 3\pi$, $- 2\pi$, $- \pi$, $0$, $\pi$ , $2\pi$, $3\pi$………
$\dfrac{{5\theta }}{2} =$…………… $n\pi$, when n is integer
$\theta = \dfrac{{2n\pi }}{5}$
Solving for $\dfrac{{\sin \theta }}{2} = 0$ :
We have $\dfrac{{\sin \theta }}{2} = 0$
$\dfrac{\theta }{2} =$ $n\pi$ 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 the 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.