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

Guest Aug 5, 2015

Best Answer 

 #1
avatar+19488 
+10

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

 

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

 

heureka  Aug 5, 2015
 #1
avatar+19488 
+10
Best Answer

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

 

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

 

heureka  Aug 5, 2015

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