Question

If $\cos 3x = - 1$, where$0^\circ \leqslant x \leqslant 360^\circ $, then find$x$
(A)$60^\circ ,180^\circ ,300^\circ $
(B)$180^\circ $
(C)$60^\circ ,180^\circ $
(D)$180^\circ ,300^\circ $

Answer 1

Hint: Simplify this equation using the identity $\cos A - \cos B = - 2\sin \dfrac{{(A + B)}}{2}\sin \dfrac{{(A - B)}}{2}$. Also, use the fact that $\sin \theta = 0$only when $\theta = n\pi $ where n is any integer to get \[x = \dfrac{{(2n - 1)\pi }}{3}\] or\[x = \dfrac{{(2n + 1)\pi }}{3}\]. Find x such that $0^\circ \leqslant x \leqslant 360^\circ $.

Complete step-by-step solution:
We have a trigonometric equation $\cos 3x = - 1$ such that $0^\circ \leqslant x \leqslant 360^\circ $.
We need to solve this equation to find the values of x. That is, we will find those x for which the given equation is satisfied.
We know that $ - 1 \leqslant \cos \theta \leqslant 1$ for any $\theta $ and $\cos 180^\circ = - 1$…..(1)
Therefore, we can compare the given equation $\cos 3x = - 1$ with this fact in (1).
Thus, we get $\cos 3x = \cos 180^\circ $.
$ \Rightarrow \cos 3x - \cos 180^\circ = 0$
Here, we will use the identity $\cos A - \cos B = - 2\sin \dfrac{{(A + B)}}{2}\sin \dfrac{{(A - B)}}{2}$.
In our case,$A = 3x$and$B = 180^\circ $.
Therefore, applying the identity we get the following equation:
\[\cos 3x - \cos 180^\circ = - 2\sin \dfrac{{(3x + 180^\circ )}}{2}\sin \dfrac{{(3x - 180^\circ )}}{2}\]
This would imply that \[ - 2\sin \dfrac{{(3x + 180^\circ )}}{2}\sin \dfrac{{(3x - 180^\circ )}}{2} = 0\]
Divide by -2 on both the sides.
Then we get
\[\sin \dfrac{{(3x + 180^\circ )}}{2}\sin \dfrac{{(3x - 180^\circ )}}{2} = 0\]……(A)
This is only possible when either $\sin \dfrac{{(3x + 180^\circ )}}{2} = 0$ or $\sin \dfrac{{(3x - 180^\circ )}}{2} = 0$
We know that $\sin \theta = 0$ only when $\theta = n\pi $ where n is any integer.
Therefore, we have either $\dfrac{{3x + 180^\circ }}{2} = n\pi $ or $\dfrac{{3x - 180^\circ }}{2} = n\pi $
Since the RHS is given in terms of radians, we will write $180^\circ $ as $\pi $.
So, we have
$\dfrac{{3x + \pi }}{2} = n\pi $ or $\dfrac{{3x - \pi }}{2} = n\pi $
$ \Rightarrow $$3x + \pi = 2n\pi $ or $3x - \pi = 2n\pi $
$ \Rightarrow $$3x = 2n\pi - \pi = (2n - 1)\pi $ or $3x = 2n\pi + \pi = (2n + 1)\pi $
Now divide by 2 throughout for both the equations. We get:
\[x = \dfrac{{(2n - 1)\pi }}{3}\]or\[x = \dfrac{{(2n + 1)\pi }}{3}\]……(I)
So this implies that $\cos 3x - \cos 180^\circ = 0$ if and only if \[x = \dfrac{{(2n - 1)\pi }}{3}\] or \[x = \dfrac{{(2n + 1)\pi }}{3}\] for any integer n.
But we are given that our x should be such that $0^\circ \leqslant x \leqslant 360^\circ $.
That is x should be such that $0 \leqslant x \leqslant 2\pi $
Now,
 $
  0 \leqslant x \leqslant 2\pi \\
   \Rightarrow 0 \leqslant \dfrac{{(2n - 1)\pi }}{3} \leqslant 2\pi \\
   \Rightarrow 0 \leqslant (2n - 1)\pi \leqslant 6\pi \\
   \Rightarrow 0 \leqslant (2n - 1) \leqslant 6 \\
   \Rightarrow 1 \leqslant 2n \leqslant 7 \\
   \Rightarrow \dfrac{1}{2} \leqslant n \leqslant \dfrac{7}{2}......(2) \\
 $
Similarly, for\[x = \dfrac{{(2n + 1)\pi }}{3}\], we have
$
  0 \leqslant x \leqslant 2\pi \\
   \Rightarrow 0 \leqslant \dfrac{{(2n + 1)\pi }}{3} \leqslant 2\pi \\
   \Rightarrow 0 \leqslant (2n + 1)\pi \leqslant 6\pi \\
   \Rightarrow 0 \leqslant (2n + 1) \leqslant 6 \\
   \Rightarrow - 1 \leqslant 2n \leqslant 5 \\
   \Rightarrow \dfrac{{ - 1}}{2} \leqslant n \leqslant \dfrac{5}{2}.....(3) \\
 $
The integer values of n which satisfy (2) or (3) are 0, 1, 2, and 3.
Put these values in equation (I)
For n = 0, we have \[x = \dfrac{\pi }{3}\] or \[x = \pi \]
For n = 1, we have\[x = \dfrac{\pi }{3}\] or \[x = \pi \]
For n = 2, we have \[x = \pi \] or \[x = \dfrac{{5\pi }}{3}\]
For n = 3, we have \[x = \dfrac{{5\pi }}{3}\] or \[x = \dfrac{{7\pi }}{3}\]
Note that\[x = \dfrac{{7\pi }}{3} \Rightarrow x = 2\pi + \dfrac{\pi }{3}\].
Therefore, the condition of $0^\circ \leqslant x \leqslant 360^\circ $ is not satisfied here.
Thus, the values for which$\cos 3x = - 1$ such that $0^\circ \leqslant x \leqslant 360^\circ $are\[x = \dfrac{\pi }{3}\],\[x = \pi \], and \[x = \dfrac{{5\pi }}{3}\].
Converting the radians into degrees, we get \[x = 60^\circ \],\[x = 180^\circ \],and\[x = 300^\circ \].

Hence the correct answer is option ‘C’ $60^\circ ,180^\circ ,300^\circ $

Note: Use $\cos 180^\circ = - 1$to get the equation$\cos 3x - \cos 180^\circ = 0$ is very essential and is the key step for solving the question. The product of two real numbers, a and b, is 0 if and only if either a = 0 or b = 0. This is an important fact that we make use of in solving this question for equation (A).