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2sec(x+60)=5sec(x-20)

 Aug 11, 2016
 #1
avatar+129839 
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2sec(x+60)=5sec(x-20)

 

2 /cos (x + 60)  = 5 / cos ( x - 20)

 

2 / [ cosxcos60 - sinxsin60]  =  5 / [ cosxcos20 + sinxsin20]

 

2  [ cosxcos20 + sinxsin20]  =  5 [ cosxcos60 - sinxsin60]

 

2 [ cosx cos20 + sinxsin20]  = 5 [ (cosx)/2 - (sqrt(3)*sinx) /2 ]

 

4[cosxcos20 + sinxsin20]  = 5 [cosx - sqrt(3)*sinx]

 

sinx [ 4sin20 + 5sqrt(3)]  = cosx [ 5 - 4cos20]

 

sinx / cosx   =   [ 5 - 4cos20] /  [ 4sin20 + 5sqrt(3)]

 

tanx  = [ 5 - 4cos20] /  [ 4sin20 + 5sqrt(3)]

 

arctan (  [ 5 - 4cos20] /  [ 4sin20 + 5sqrt(3)])  = x =  about 7.056° + n*360°     where n is an integer

 

Here's a graph :  https://www.desmos.com/calculator/oqxckkdfx2

 

 

 

cool cool cool

 Aug 11, 2016
 #2
avatar+26388 
0

2sec(x+60)=5sec(x-20)

 

\(\begin{array}{|rcll|} \hline 2\cdot \sec(x+60^{\circ}) &=& 5\cdot \sec(x-20^{\circ}) \\ \text{We substitute } x = y-20^{\circ} \\\\ 2\cdot \sec(y-20^{\circ}+60^{\circ}) &=& 5\cdot \sec(y-20^{\circ}-20^{\circ}) \\ 2\cdot \sec(y+40^{\circ}) &=& 5\cdot \sec(y-40^{\circ}) \quad | \quad \sec(\phi) = \frac{1}{\cos(\phi)}\\ \frac{ 2 } { \cos(y+40^{\circ}) } &=& \frac{ 5 } { \cos(y-40^{\circ}) } \\ \frac{ \cos(y-40^{\circ}) } { \cos(y+40^{\circ}) } &=& \frac{ 5 } { 2 } \quad | \quad \cos(y-40^{\circ}) = \cos(y)\cos(40^{\circ})+\sin(y)\sin(40^{\circ}) \\ \frac{ \cos(y)\cos(40^{\circ})+\sin(y)\sin(40^{\circ}) } { \cos(y+40^{\circ}) } &=& \frac{ 5 } { 2 } \quad | \quad \cos(y+40^{\circ}) = \cos(y)\cos(40^{\circ})-\sin(y)\sin(40^{\circ}) \\ \frac{ \frac{\cos(y)\cos(40^{\circ})+\sin(y)\sin(40^{\circ})} {\cos(y)\cos(40^{\circ})} } { \frac{\cos(y)\cos(40^{\circ})-\sin(y)\sin(40^{\circ})} {\cos(y)\cos(40^{\circ})} } &=& \frac{ 5 } { 2 } \\ \frac{ 1+\tan(y)\tan(40^{\circ}) } {1-\tan(y)\tan(40^{\circ}) } &=& \frac{ 5 } { 2 } \\\\ 2\cdot [~ 1+\tan(y)\tan(40^{\circ}) ~] &=& 5\cdot [~ 1-\tan(y)\tan(40^{\circ}) ~] \\ 2 + 2\cdot \tan(y)\tan(40^{\circ}) &=& 5 - 5 \cdot \tan(y)\tan(40^{\circ}) \\ 7\cdot \tan(y)\tan(40^{\circ}) &=& 3 \\ \tan(y) &=& \frac37\cdot \frac{1}{\tan(40^{\circ})} \quad | \quad y=x+20^{\circ}\\ \tan(x+20^{\circ}) &=& \frac37\cdot \frac{1}{\tan(40^{\circ})} \\ x+20^{\circ} &=& \arctan{\left( \frac37\cdot \frac{1}{\tan(40^{\circ})} \right)} \pm n\cdot 180^{\circ}\\ x &=& -20^{\circ} + \arctan{\left( \frac37\cdot \frac{1}{\tan(40^{\circ})} \right)} \pm n\cdot 180^{\circ}\\ x &=& -20^{\circ} + \arctan{\left( 0.51075153968 \right)} \pm n\cdot 180^{\circ}\\ x &=& -20^{\circ} + 27.0557431286^{\circ} \pm n\cdot 180^{\circ}\\\\ \mathbf{x} &\mathbf{=}& \mathbf{7.0557431286^{\circ} \pm n\cdot 180^{\circ} } \qquad n \in N\\ \hline \end{array}\)

 

laugh

 Aug 12, 2016
edited by heureka  Aug 12, 2016

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