Question

The general solution of the equation $\cos x . \cos 6x = - 1$, is :

• A. x = (2n + 1)π, n ∈ I
• B. x = 2nπ, n ∈ I
• C. x = (2n - 1)π, n ∈ I
• D. none of these

Answer 1

Answer: (A)

x = (2n + 1)π, n ∈ I

Answer 2

The equation given is $\cos x \cdot \cos 6x = -1$.

We know that the range of the cosine function is $[-1, 1]$, i.e., $-1 \le \cos \theta \le 1$.

For the product of two cosine terms to be $-1$, there are only two possible scenarios:

Scenario 1: $\cos x = 1$ and $\cos 6x = -1$

If $\cos x = 1$, then $x = 2k\pi$ for some integer $k$.

Substitute this value of $x$ into the second condition: $\cos(6 \cdot 2k\pi) = -1$ $\cos(12k\pi) = -1$

However, $12k$ is always an even integer. We know that $\cos(m\pi) = (-1)^m$. If $m$ is even, $\cos(m\pi) = 1$.

So, $\cos(12k\pi) = 1$.

This leads to $1 = -1$, which is a contradiction. Therefore, Scenario 1 yields no solution.

Scenario 2: $\cos x = -1$ and $\cos 6x = 1$

If $\cos x = -1$, then $x = (2n+1)\pi$ for some integer $n$.

Substitute this value of $x$ into the second condition: $\cos(6 \cdot (2n+1)\pi) = 1$ $\cos((12n+6)\pi) = 1$

Here, $(12n+6)$ is always an even integer for any integer $n$ (since $12n$ is even and $6$ is even, their sum is even).

As established in Scenario 1, if $m$ is an even integer, $\cos(m\pi) = 1$.

So, $\cos((12n+6)\pi) = 1$.

This condition is satisfied.

Therefore, the general solution for the given equation is $x = (2n+1)\pi$, where $n \in I$.