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Question: Find the general solution of sin2x + cosx = 0...

Find the general solution of sin2x + cosx = 0

Explanation

Solution

Hint: First we will use the trigonometric formula sin2x=2sinxcosx\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 sin2x=2sinxcosx\sin 2x=2\sin x\cos x , to expand sin2x and then we will take cosx common.
2sinxcosx+cosx=0 cosx(2sinx+1)=0 \begin{aligned} & 2\sin x\cos x+\cos x=0 \\\ & \cos x\left( 2\sin x+1 \right)=0 \\\ \end{aligned}
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 cosπ2\cos \dfrac{\pi }{2} = 0,
Hence, we can say that cosx = cosπ2\cos \dfrac{\pi }{2}.
Now we will use the formula for general solution of cos,
Now, if we have cosθ=cosα\cos \theta =\cos \alpha then the general solution is:
θ=2nπ±α\theta =2n\pi \pm \alpha
Now using the above formula for cosx = cosπ2\cos \dfrac{\pi }{2} we get,
x=2nπ±π2............(1)x=2n\pi \pm \dfrac{\pi }{2}............(1)
Here n = integer.
Now we will find the general solution of 2sinx + 1 = 0
sinx=12 sinx=sinπ6 \begin{aligned} & \sin x=\dfrac{-1}{2} \\\ & \sin x=\sin \dfrac{-\pi }{6} \\\ \end{aligned}
Now we will use the formula for general solution of sin,
Now, if we have sinθ=sinα\sin \theta =\sin \alpha then the general solution is:
θ=nπ+(1)nα\theta =n\pi +{{\left( -1 \right)}^{n}}\alpha
Now using the above formula for sinx=sinπ6\sin x=\sin \dfrac{-\pi }{6} we get,
x=nπ+(1)n(π6).............(2)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=2nπ±π2 or x=nπ+(1)n(π6)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 sin2x=2sinxcosx\sin 2x=2\sin x\cos x that we have used must be kept in mind. One can also take some different value of α\alpha like in cosx = 0 we can take 3π2\dfrac{3\pi }{2} instead of π2\dfrac{\pi }{2} , and then can apply the same formula for the general solution and the answer that we get will also be correct.