How do you simplify $sin 2 x + cos x = 0$ $sin2x + cosx = 0$ using the double angle identity?

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How do you simplify $sin 2 x + cos x = 0$ $sin2x + cosx = 0$ using the double angle identity?

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Answer 1 · 2018-04-26T12:50:32

General solution is $x = (2 n + 1) \frac{π}{2}$ $x=(2n+1)pi/2$ or $x = n π \pm {(- 1)}^{n} (- \frac{π}{6})$ $x=npi+-(-1)^n(-pi/6)$ , where $n$ $n$ is an integer.
In the interval $[0, 2 π)$ $[0,2pi)$ , $x \in {\frac{π}{2}, \frac{7 π}{6}, \frac{3 p}{2}, \frac{11 π}{6}}$ $x in{pi/2,(7pi)/6,(3p)/2,(11pi)/6}$

Explanation:

As $sin 2 x = 2 sin x cos x$ $sin2x=2sinxcosx$ , we can write $sin 2 x + cos x = 0$ $sin2x+cosx=0$ as

$2 sin x cos x + cos x = 0$ $2sinxcosx+cosx=0$

or $cos x (2 sin x + 1) = 0$ $cosx(2sinx+1)=0$

Hence either $cos x = 0$ $cosx=0$ or $2 sin x + 1 = 0$ $2sinx+1=0$ i.e. $sin x = - \frac{1}{2}$ $sinx=-1/2$

If $cos x = 0$ $cosx=0$ , $x = (2 n + 1) \frac{π}{2}$ $x=(2n+1)pi/2$ , where $n$ $n$ is an integer

and if $sin x = - \frac{1}{2} = sin (- \frac{π}{6})$ $sinx=-1/2=sin(-pi/6)$ , $x = n π \pm {(- 1)}^{n} (- \frac{π}{6})$ $x=npi+-(-1)^n(-pi/6)$

in the interval $[0, 2 π)$ $[0,2pi)$ ,

$x \in {\frac{π}{2}, \frac{7 π}{6}, \frac{3 p}{2}, \frac{11 π}{6}}$ $x in{pi/2,(7pi)/6,(3p)/2,(11pi)/6}$

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How do you simplify $sin 2 x + cos x = 0$ $sin2x + cosx = 0$ using the double angle identity?

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