Question

The solution of the equation `sin 7x + cos2x = -2` is/are?

A) `x = 2kpi/7 + 3pi/14, k in I`
B)`x = npi + pi/4, n in I`
C)`x=2npi + pi/2, n in I`
D)`none of these`

Answer 1

`C`

Explanation:

Since both the sine and cosine functions have a minimum of `-1`, to obtain `-2` as their sum we need to minimize both functions with the same value of `x`.

Moreover, we know that `sin((3pi)/2 + 2k\pi)=-1`, and `cos(\pi + 2k\pi)=-1`

At this point, you should be able to see that the correct answer is `C`. If you consider the angle `x=pi/2` (periodicity apart), you can see that

`sin(7x) = sin((7pi)/2) = sin((3pi)/2 + (4pi)/2) = sin((3pi)/2 + 2\pi) =`

`= sin((3pi)/2) = -1`

For the same reason, we have

`cos(2x) = cos(\pi) = -1`

And finally, `sin(7x)+cos(2x)=-1+(-1) = -1-1=-2`

Since `\pi/2` is a good angle, all the angles of the form `pi/2+2k\pi` are good as well, thus justifying the answer.