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I am supposed to find all of the solutions to the equation below in the interval [0, 360). Round approximate answers to the nearest tenth of a degree. 

The equation is: sin(x) - cos(x) = 0.8

 

What's the most efficient way to solve this? 

 Apr 30, 2019
 #1
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I am supposed to find all of the solutions to the equation below in the interval [0, 360).

Round approximate answers to the nearest tenth of a degree. 

The equation is: sin(x) - cos(x) = 0.8

What's the most efficient way to solve this? 

 

I.

Formula:

\(\begin{array}{|rcll|} \hline \sin(x-45^\circ) &=& \sin(x)\cos(45^\circ) - \cos(x)\sin(45^\circ) \\\\ && \qquad \boxed{\sin(45^\circ)=\cos(45^\circ)=\dfrac{\sqrt{2}} {2}} \\\\ \sin(x-45^\circ) &=& \sin(x)\dfrac{\sqrt{2}} {2} - \cos(x)\dfrac{\sqrt{2}} {2} \\\\ \sin(x-45^\circ) &=& \dfrac{\sqrt{2}} {2} \Big(\sin(x) - \cos(x) \Big) \\\\ \mathbf{\sin(x) - \cos(x)} &\mathbf{=}& \mathbf{\dfrac{2} {\sqrt{2}}\sin(x-45^\circ)} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \sin(x) - \cos(x) &=& 0.8 \quad | \quad \mathbf{\sin(x) - \cos(x) = \dfrac{2} {\sqrt{2}}\sin(x-45^\circ)} \\ \dfrac{2} {\sqrt{2}}\sin(x-45^\circ) &=& 0.8 \\ \sin(x-45^\circ) &=& 0.8 \dfrac{\sqrt{2}}{2} \\ \mathbf{\sin(x-45^\circ)} &\mathbf{=}& \mathbf{0.4\sqrt{2}} \\ \hline \end{array} \)

 

Solution 1:

\(\begin{array}{|rcll|} \hline \mathbf{\sin(x-45^\circ)} &\mathbf{=}& \mathbf{0.4\sqrt{2}} \quad | \quad \arcsin() \\ x-45^\circ & = & \arcsin(0.4\sqrt{2})+ 360z,\ z \in \mathbb{Z} \\ x & = & 45^\circ + \arcsin(0.4\sqrt{2})+ 360z \\ & = & 45^\circ + \arcsin(0.56568542495)+ 360z \\ & = & 45^\circ + 34.4499019880^\circ + 360z \\ & = & 79.4499019880^\circ + 360z \\ & = & 79.4499019880^\circ \quad | \quad \text{interval [0, 360)} \\ \mathbf{ x } & \mathbf{=} & \mathbf{79.4^\circ} \\ \hline \end{array} \)

 

Solution 2:

\(\begin{array}{|rcll|} \hline \mathbf{\sin(x-45^\circ)} &\mathbf{=}& \mathbf{0.4\sqrt{2}} \\\\ && \boxed{\sin(x-45^\circ) = \sin\Big(180^\circ-(x-45^\circ)\Big) \\ \sin(x-45^\circ) = \sin(180^\circ-x+45^\circ) \\ \mathbf{\sin(x-45^\circ) = \sin(225^\circ-x)} } \\\\ \mathbf{\sin(225^\circ-x)} &\mathbf{=}& \mathbf{0.4\sqrt{2}} \quad | \quad \arcsin() \\ 225^\circ-x & = & \arcsin(0.4\sqrt{2})+ 360z,\ z \in \mathbb{Z} \\ x &=& 225^\circ - \arcsin(0.4\sqrt{2})+ 360z \\ &=& 225^\circ - \arcsin(0.56568542495)+ 360z \\ &=& 225^\circ - 34.4499019880^\circ + 360z \\ &=& 190.550098012^\circ + 360z \quad | \quad \text{interval [0, 360)} \\ &=& 190.550098012^\circ \\ \mathbf{ x } & \mathbf{=} & \mathbf{190.6^\circ} \\ \hline \end{array}\)

 

laugh

 Apr 30, 2019

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