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# 2cos^2(x)-sin(x)-1=0

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2cos^2(x)-sin(x)-1=0

Guest Aug 5, 2015

#1
+18827
+10

2cos^2(x)-sin(x)-1=0

$$\small{\text{ \begin{array}{rclr} 2\cos^2{(x)}-\sin{(x)}-1 &=& 0 \qquad | \qquad \cos^2{(x)}=1-\sin^2{(x)} \\ 2 [ 1-\sin^2{(x)} ]-\sin{(x)}-1 &=& 0 \\ 2 - 2 \sin^2{(x)} -\sin{(x)}-1 &=& 0 \\ - 2 \sin^2{(x)} -\sin{(x)}+1 &=& 0 \qquad | \qquad \cdot (-1)\\ 2 \sin^2{(x)} +\sin{(x)}-1 &=& 0 \\ \left[ 2 \sin{(x)} -1 \right]\cdot \left[ \sin{(x)}+1 \right] &=& 0 \\\\ \underbrace{\left[ {2 \sin{(x)} -1}\right]}_{=0} \cdot \underbrace{\left[ \sin{(x)}+1 \right]}_{=0} &=& 0 \\\\ \sin{(x)}+1 &=& 0 \\ \sin{(x)} &=& -1\\ x &=& \arcsin(-1) \pm 2k\pi \\ \mathbf{x} & \mathbf{=} & \mathbf{-\dfrac{\pi}{2} \pm 2k\pi } \quad \mathbf{k}\in Z\\ \\ \hline \\ {2 \sin{(x)} -1} &{=}&{0} \\ 2 \sin{(x)} &=&1 \\\\ \sin{(x)} &=&\dfrac{1}{2} \\ x &=& \arcsin\left(\dfrac{1}{2}\right) \pm 2k\pi \\ \mathbf{x} & \mathbf{=} & \mathbf{\dfrac{\pi} {6} \pm 2k\pi} \mathbf \quad {k}\in Z\\\\ \sin{(\pi-x)} &=&\dfrac{1}{2} \\ \pi-x &=& \arcsin\left(\dfrac{1}{2}\right) \pm 2k\pi \\ \pi-x &=& \dfrac{\pi} {6} \pm 2k\pi \\\\ \mathbf{x} & \mathbf{=} & \mathbf{\dfrac{5} {6}\pi \pm 2k\pi} \quad \mathbf{k}\in Z \end{array} }}$$

heureka  Aug 5, 2015
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#1
+18827
+10

2cos^2(x)-sin(x)-1=0

$$\small{\text{ \begin{array}{rclr} 2\cos^2{(x)}-\sin{(x)}-1 &=& 0 \qquad | \qquad \cos^2{(x)}=1-\sin^2{(x)} \\ 2 [ 1-\sin^2{(x)} ]-\sin{(x)}-1 &=& 0 \\ 2 - 2 \sin^2{(x)} -\sin{(x)}-1 &=& 0 \\ - 2 \sin^2{(x)} -\sin{(x)}+1 &=& 0 \qquad | \qquad \cdot (-1)\\ 2 \sin^2{(x)} +\sin{(x)}-1 &=& 0 \\ \left[ 2 \sin{(x)} -1 \right]\cdot \left[ \sin{(x)}+1 \right] &=& 0 \\\\ \underbrace{\left[ {2 \sin{(x)} -1}\right]}_{=0} \cdot \underbrace{\left[ \sin{(x)}+1 \right]}_{=0} &=& 0 \\\\ \sin{(x)}+1 &=& 0 \\ \sin{(x)} &=& -1\\ x &=& \arcsin(-1) \pm 2k\pi \\ \mathbf{x} & \mathbf{=} & \mathbf{-\dfrac{\pi}{2} \pm 2k\pi } \quad \mathbf{k}\in Z\\ \\ \hline \\ {2 \sin{(x)} -1} &{=}&{0} \\ 2 \sin{(x)} &=&1 \\\\ \sin{(x)} &=&\dfrac{1}{2} \\ x &=& \arcsin\left(\dfrac{1}{2}\right) \pm 2k\pi \\ \mathbf{x} & \mathbf{=} & \mathbf{\dfrac{\pi} {6} \pm 2k\pi} \mathbf \quad {k}\in Z\\\\ \sin{(\pi-x)} &=&\dfrac{1}{2} \\ \pi-x &=& \arcsin\left(\dfrac{1}{2}\right) \pm 2k\pi \\ \pi-x &=& \dfrac{\pi} {6} \pm 2k\pi \\\\ \mathbf{x} & \mathbf{=} & \mathbf{\dfrac{5} {6}\pi \pm 2k\pi} \quad \mathbf{k}\in Z \end{array} }}$$

heureka  Aug 5, 2015

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