Question

How do you solve for x in $\sin (2x) + \cos (x) = 0$?

Answer 1

Sine and cosine are two most basic trigonometric functions. Sine is the ratio of height and hypotenuse and cosine is the ratio of base and the hypotenuse. These two can be interconverted very easily and this property will be useful for us in this question.

Complete step by step solution:
According to the question we have to solve for x in the equation $\sin (2x) + \cos (x) = 0$
So, we have to use trigonometric formulas to simplify this equation. We can either convert a cosine into a sine function or a sine function into a cosine function. But in the question, we can clearly observe that the angle of the sine function is larger than the angle of the cosine function. So we will prefer converting the sine function into a cosine function.
To do this, we have to use this trigonometric formula
$\Rightarrow \sin 2x = 2\sin (x)\cos (x)$
Now we have to just solve it
$\Rightarrow 2\sin x\cos x + \cos x = 0$
$\Rightarrow \cos x(2\sin x + 1) = 0$
Now, in this case, we have two situations, either $\cos x$ is zero or either $2\sin x + 1$ is zero.
So we have to solve it one by one
First let us consider $\cos x$ is zero
$\cos x = 0$, it means that we have to consider the general solution of $\cos x$
So, $x = (2n + 1)\dfrac{\pi }{2}$, where $n \in Z$, it means all odd multiples of $\dfrac{\pi }{2}$.
This general solution will give all answers when $\cos x$ is zero
Now let us consider $2\sin x + 1$ is zero
$\Rightarrow 2\sin x + 1 = 0$
$\Rightarrow 2\sin x = - 1$
$\Rightarrow \sin x = \dfrac{{ - 1}}{2}$
We know that $\sin \dfrac{{7\pi }}{6} = \dfrac{{ - 1}}{2}$
So, let us say$\sin \dfrac{{7\pi }}{6} = \sin \theta$, so$\theta = \dfrac{{7\pi }}{6}$, it means that we have to consider the general solution of $\sin x$
So, $x = n\pi + {( - 1)^2}\dfrac{{7\pi }}{6}$, where $n \in Z$
This general solution will give all answers when$2\sin x + 1$ is zero

Hence, our solution is either $x = (2n + 1)\dfrac{\pi }{2}$ or $x = n\pi + {( - 1)^2}\dfrac{{7\pi }}{6}$.

Note:
We have to remember the general solution of these equations otherwise, we will only give one solution and it will be wrong. We can observe that there are infinite solutions. Using the formula of converting the sine function into a cosine function is very important.