How do you solve $\sin 2x-\cos x=0$?
Answer
581.7k+ views
Hint: We will use the double angle formula for the sine function, which is given as $\sin 2x=2\sin x\cos x$. Then we will take the cosine function common. After that we will get two factors of the given equation such that their product is zero. Then we will equate these factors to zero and find the possible values of the variable $x$.
Complete step by step answer:
The given equation is $\sin 2x-\cos x=0$. We know that there is a double angle formula for the sine function. This formula is given as $\sin 2x=2\sin x\cos x$. Substituting this formula in the given equation, we get the following expression,
$2\sin x\cos x-\cos x=0$
Both the terms on the left hand side have $\cos x$. We will take out $\cos x$ as a common factor. SO, we get the following equation,
$\cos x\left( 2\sin x-1 \right)=0$
Now, we have obtained two factors of the given expression such that their product is 0. We know that if the product of two numbers is 0, then either one of the two numbers has to be 0. Therefore, we have two possibilities, that is, either one of the factors in the above equation can be 0. So, we have either $\cos x=0$ or $2\sin x-1=0$.
If $\cos x=0$, then we know that $x=\dfrac{\pi }{2}$. The cosine function is a periodic function with a period of $2\pi $. Therefore, we have $x=\left( 2n+1 \right)\dfrac{\pi }{2}$ where $n$ is a natural number.
If $2\sin x-1=0$, we have $\sin x=\dfrac{1}{2}$. We know that if $\sin x=\dfrac{1}{2}$, then $x=\dfrac{\pi }{6}$. The sine function is a periodic function and its period is $2\pi $. Therefore, we have $x=\dfrac{\pi }{6}+2\pi n$ where $n$ is a natural number.
Therefore, the solution of the given equation is $x=\left( 2n+1 \right)\dfrac{\pi }{2}\text{ or }x=\dfrac{\pi }{6}+2\pi n$ where $n$ is a natural number.
Note:
We should be familiar with the values of the trigonometric functions for standard angles. These values are useful in such types of questions. The double angle formulae and the half angle formulae are also essential in simplification of an equation. An important aspect of the trigonometric functions is their periodicity. A periodic function is a function that repeats its values after a fixed period.
Complete step by step answer:
The given equation is $\sin 2x-\cos x=0$. We know that there is a double angle formula for the sine function. This formula is given as $\sin 2x=2\sin x\cos x$. Substituting this formula in the given equation, we get the following expression,
$2\sin x\cos x-\cos x=0$
Both the terms on the left hand side have $\cos x$. We will take out $\cos x$ as a common factor. SO, we get the following equation,
$\cos x\left( 2\sin x-1 \right)=0$
Now, we have obtained two factors of the given expression such that their product is 0. We know that if the product of two numbers is 0, then either one of the two numbers has to be 0. Therefore, we have two possibilities, that is, either one of the factors in the above equation can be 0. So, we have either $\cos x=0$ or $2\sin x-1=0$.
If $\cos x=0$, then we know that $x=\dfrac{\pi }{2}$. The cosine function is a periodic function with a period of $2\pi $. Therefore, we have $x=\left( 2n+1 \right)\dfrac{\pi }{2}$ where $n$ is a natural number.
If $2\sin x-1=0$, we have $\sin x=\dfrac{1}{2}$. We know that if $\sin x=\dfrac{1}{2}$, then $x=\dfrac{\pi }{6}$. The sine function is a periodic function and its period is $2\pi $. Therefore, we have $x=\dfrac{\pi }{6}+2\pi n$ where $n$ is a natural number.
Therefore, the solution of the given equation is $x=\left( 2n+1 \right)\dfrac{\pi }{2}\text{ or }x=\dfrac{\pi }{6}+2\pi n$ where $n$ is a natural number.
Note:
We should be familiar with the values of the trigonometric functions for standard angles. These values are useful in such types of questions. The double angle formulae and the half angle formulae are also essential in simplification of an equation. An important aspect of the trigonometric functions is their periodicity. A periodic function is a function that repeats its values after a fixed period.
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