
Find the general solution of sin2x + cosx = 0
Answer
596.7k+ views
Hint: First we will use the trigonometric formula $\sin 2x=2\sin x\cos x$ , and then we will take the term cosx common and then use the formula for finding the general solution of two different equations and that will be the answer.
Complete step-by-step answer:
Let’s start solving the question,
sin2x + cosx = 0
Now we will use $\sin 2x=2\sin x\cos x$ , to expand sin2x and then we will take cosx common.
$\begin{align}
& 2\sin x\cos x+\cos x=0 \\
& \cos x\left( 2\sin x+1 \right)=0 \\
\end{align}$
Now we have converted it into two equation and we will solve it separately,
cosx = 0 and 2sinx + 1 = 0
Let’s first solve cosx = 0,
We know that $\cos \dfrac{\pi }{2}$ = 0,
Hence, we can say that cosx = $\cos \dfrac{\pi }{2}$.
Now we will use the formula for general solution of cos,
Now, if we have $\cos \theta =\cos \alpha $ then the general solution is:
$\theta =2n\pi \pm \alpha $
Now using the above formula for cosx = $\cos \dfrac{\pi }{2}$ we get,
$x=2n\pi \pm \dfrac{\pi }{2}............(1)$
Here n = integer.
Now we will find the general solution of 2sinx + 1 = 0
$\begin{align}
& \sin x=\dfrac{-1}{2} \\
& \sin x=\sin \dfrac{-\pi }{6} \\
\end{align}$
Now we will use the formula for general solution of sin,
Now, if we have $\sin \theta =\sin \alpha $ then the general solution is:
$\theta =n\pi +{{\left( -1 \right)}^{n}}\alpha $
Now using the above formula for $\sin x=\sin \dfrac{-\pi }{6}$ we get,
$x=n\pi +{{\left( -1 \right)}^{n}}\left( \dfrac{-\pi }{6} \right).............(2)$
Here n = integer.
Now from equation (1) and (2) we can say that the answer is,
$x=2n\pi \pm \dfrac{\pi }{2}\text{ or }x=n\pi +{{\left( -1 \right)}^{n}}\left( \dfrac{-\pi }{6} \right)$
Hence, this is the answer to this question.
Note: The trigonometric formula $\sin 2x=2\sin x\cos x$ that we have used must be kept in mind. One can also take some different value of $\alpha $ like in cosx = 0 we can take $\dfrac{3\pi }{2}$ instead of $\dfrac{\pi }{2}$ , and then can apply the same formula for the general solution and the answer that we get will also be correct.
Complete step-by-step answer:
Let’s start solving the question,
sin2x + cosx = 0
Now we will use $\sin 2x=2\sin x\cos x$ , to expand sin2x and then we will take cosx common.
$\begin{align}
& 2\sin x\cos x+\cos x=0 \\
& \cos x\left( 2\sin x+1 \right)=0 \\
\end{align}$
Now we have converted it into two equation and we will solve it separately,
cosx = 0 and 2sinx + 1 = 0
Let’s first solve cosx = 0,
We know that $\cos \dfrac{\pi }{2}$ = 0,
Hence, we can say that cosx = $\cos \dfrac{\pi }{2}$.
Now we will use the formula for general solution of cos,
Now, if we have $\cos \theta =\cos \alpha $ then the general solution is:
$\theta =2n\pi \pm \alpha $
Now using the above formula for cosx = $\cos \dfrac{\pi }{2}$ we get,
$x=2n\pi \pm \dfrac{\pi }{2}............(1)$
Here n = integer.
Now we will find the general solution of 2sinx + 1 = 0
$\begin{align}
& \sin x=\dfrac{-1}{2} \\
& \sin x=\sin \dfrac{-\pi }{6} \\
\end{align}$
Now we will use the formula for general solution of sin,
Now, if we have $\sin \theta =\sin \alpha $ then the general solution is:
$\theta =n\pi +{{\left( -1 \right)}^{n}}\alpha $
Now using the above formula for $\sin x=\sin \dfrac{-\pi }{6}$ we get,
$x=n\pi +{{\left( -1 \right)}^{n}}\left( \dfrac{-\pi }{6} \right).............(2)$
Here n = integer.
Now from equation (1) and (2) we can say that the answer is,
$x=2n\pi \pm \dfrac{\pi }{2}\text{ or }x=n\pi +{{\left( -1 \right)}^{n}}\left( \dfrac{-\pi }{6} \right)$
Hence, this is the answer to this question.
Note: The trigonometric formula $\sin 2x=2\sin x\cos x$ that we have used must be kept in mind. One can also take some different value of $\alpha $ like in cosx = 0 we can take $\dfrac{3\pi }{2}$ instead of $\dfrac{\pi }{2}$ , and then can apply the same formula for the general solution and the answer that we get will also be correct.
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