Question

How do you solve $\sin x + \cos x = 0$?

Answer 1

Hint:We know that the value of x remains between ${0^ \circ }$ and ${360^ \circ }$. The domain of both $\sin x$ and $\cos x$ is $\left( { - \infty ,\infty } \right)$ and range is between $\left[ { - 1,1} \right]$. Therefore, the domain of $\sin x + \cos x$ is $\left( { - \infty ,\infty } \right)$. The range of $\sin x + \cos x$ is between $\left[ { - \sqrt {2,} \sqrt 2 } \right]$ which can be easily found out by differentiating $\sin x + \cos x$.

Complete step by step solution:
Given is, $\sin x + \cos x = 0$
Taking $\cos x$ on right hand side and leaving $\sin x$ alone on left side gives us,
$\sin x = - \cos x$
Now, dividing the entire equation by $ - \cos x$ we get;
$\dfrac{{\sin x}}{{ - \cos x}} = \dfrac{{ - \cos x}}{{ - \cos x}}$
$ - \tan x = 1$
In this next step we have to shift minus (-) sign from left hand side to right hand side so that we get;
$\tan x = - 1$
Now, taking the inverse tangent we get;
$x = {\tan ^{ - 1}}\left( { - 1} \right)$
We, know that $\tan x = 1$ at $\dfrac{\pi }{4}$
so, ${\tan ^{ - 1}}\left( { - 1} \right) = - \dfrac{\pi }{4}$
This brings us to the value of x being $ - \dfrac{\pi }{4}$
$x = - \dfrac{\pi }{4}$
We are aware that the tangent function is negative in the second and fourth quadrants. To find the second solution, we have to subtract the obtained angle from π to find the solution in the third quadrant.
$x = - \dfrac{\pi }{4} - \pi $
Now we need to simply the equation,
$x = - \dfrac{{5\pi }}{4}$, now adding $2\pi $ to $\dfrac{{ - 5\pi }}{4}$
$x = \dfrac{{ - 5\pi }}{4} + 2\pi $
$x = \dfrac{{3\pi }}{4}$
Now, for the last step we need to find the period of the function. The period of a function can be calculated using $\dfrac{\pi }{{|b|}}$. We have to replace b with $1$ in the given formula in order to find the period for our function.
$\dfrac{\pi }{{|b|}} = \dfrac{\pi }{1} = \pi $
To check whether every negative angle gives out positive angle we add $\pi $ to $\dfrac{{ - \pi
}}{4}$.
$\dfrac{{ - \pi }}{4} + \pi = \dfrac{{3\pi }}{4}$
The period of the $\tan x$ function is $\pi $ so values will repeat every $\pi $$\sin x + \cos x = 0$radians in both directions.
Hence the answer is $x = \dfrac{{3\pi }}{4} + \pi n,\dfrac{{3\pi }}{4} + \pi n$, for any integer $n$

Note: In the first quadrant, both $\sin x$ and $\cos x$ are non-negative and hence their sum cannot be zero. $\cos x$ is zero at multiples of 90 degrees and $\sin x$ is zero at 0 degrees and multiples of 180 degrees, so they are never zero for the same x.
Solutions to $\sin x + \cos x = 0$ require they both be of opposite signs, which occurs in quadrants 2 and 4. Also, their numerical magnitudes must be equal so that they cancel and get zero.