Question

How do you find the exact solution of the equation $\sin 2x+\cos x=0$ in the interval $[0,2\pi )$?

Answer 1

We are given an equation and we have to find its solution in the given interval $[0,2\pi )$. We will first expand the term $\sin 2x$ as $2\sin x\cos x$. Then, we will take $\cos x$ out as common. So, we are then left with the following expression, $\cos x(2\sin x+1)=0$. Equating each of the terms to zero, we will have the solution of x for each. Hence, we will get the exact solution of the given equation.

Complete step by step solution:
According to the given question, we are given an equation $\sin 2x+\cos x=0$ and we have to find the exact solution in the given interval $[0,2\pi )$.
We will begin by writing the given equation, we have,
$\sin 2x+\cos x=0$----(1)
We will now expand the term $\sin 2x$ as $2\sin x\cos x$, so we will get the expression as,
$\Rightarrow 2\sin x\cos x+\cos x=0$----(2)
As we can see that in the above expression $\cos x$ is common in both the terms, so we can take $\cos x$ out as common from the terms and we will have the expression as,
$\Rightarrow \cos x(2\sin x+1)=0$----(3)
Equating each of the terms with zero, we will have the following expression,
Either $\cos x=0$ or $2\sin x+1=0$
When $\cos x=0$, x can have the following values, which are,
$x=\dfrac{\pi }{2},\dfrac{3\pi }{2}$
When $2\sin x+1=0$, then we have,
Subtracting 1 on either side of the above expression, we have,
$\Rightarrow 2\sin x=-1$
$\Rightarrow \sin x=\dfrac{-1}{2}$
Now, x can have the following values, which are,
$x=\dfrac{7\pi }{6},\dfrac{11\pi }{6}$
Since, we are given that the solution should be in the interval $[0,2\pi )$,
Therefore, the values of $x=\dfrac{\pi }{2},\dfrac{3\pi }{2},\dfrac{7\pi }{6},\dfrac{11\pi }{6}$

Note: The equation should be carefully equated. And the values of the trigonometric functions should be correctly known and written. While writing the final solution of the given equation, the given interval should be kept under consideration.