We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
108
1
avatar

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
avatar+23071 
+3

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

23 Online Users

avatar
avatar
avatar
avatar
avatar