Question

If $\cos 3x = - 1$ when $0 \geqslant x \geqslant 360^\circ $then $x$
A. $60^\circ ,180^\circ ,300^\circ $
B. $180^\circ $
C. $60^\circ ,180^\circ $
D. $180^\circ ,300^\circ $

Answer 1

Hint:
We know that for $\cos \theta = - 1$ the general solution will be $\theta = (2n + 1)\pi $ where n is the whole number and $n \in 0, 1, 2, 3........$
Here we have $\theta = 3x$ and now by putting the value of $n$ we get the suitable $x$.

Complete step by step solution:
Here we are given the equation that $\cos 3x = - 1$
And we know that $\cos 180^\circ = - 1$
So we can write that $\cos \pi = - 1$
So $\cos 3x = \cos \pi $
Now we can use the formula that if$\cos \theta = \cos \alpha $, then $\theta = 2n\pi \pm \alpha $
So here we are given that $\theta = 3x,\alpha = \pi $
So we get that
$3x = 2n\pi \pm \pi $
This is the general solution.
Now we get that $x = (2n \pm 1)\dfrac{\pi }{3}$
Here n is the whole number $n \in 0,1,2,3........$
Here $2n \pm 1$ is the odd multiple of $\pi $ therefore we can also write that
$x = (2n + 1)\dfrac{\pi }{3}$ and where n belongs to the integer
Which means that $n = (0, \pm 1, \pm 2, \pm 3,..........)$
Now we are given that $0 \geqslant x \geqslant 360^\circ $
So we can say that $x = (2n + 1)\dfrac{\pi }{3}$
If $n = 0,x = \dfrac{\pi }{3},x = 60^\circ $
If $n = - 1,x = - \dfrac{\pi }{3},x = - 60^\circ $
So for n is negative we get $x = - 60^\circ $ therefore n cannot be negative
If $n = 1,x = \pi ,x = 180^\circ $
If $n = 2,x = \dfrac{{5\pi }}{3},x = 300^\circ $
If $n = 3,x = \dfrac{{7\pi }}{3},x = 420^\circ $
As we can see that $420^\circ $ is not in the range $0 \geqslant x \geqslant 360^\circ $

Therefore $x$ must be $60^\circ ,180^\circ ,300^\circ $ for $\cos 3x = - 1$

Note:
If we are given that $\cos \theta = \cos \alpha $ then its general solution will be given as $\theta = 2n\pi \pm \alpha $
Similarly for $\sin \theta = \sin \alpha $ its general solution is given as $\theta = n\pi + {( - 1)^n}\alpha $ and for $\tan \theta = \tan \alpha $ we have the general formula as $\theta = n\pi + \alpha $.