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Solve 2cos^2θ/2 = cos2θ for 0º≤θ≤360º. (Hint: Graph each side of the equation and find the points of intersection.)

A) ≈73.8º; 260.4º

B) ≈141.3º; 218.7º

C) ≈225.7º; 315.8º

D) ≈51.7º; 138.2º

Guest Oct 22, 2014

Best Answer 

 #2
avatar+81055 
+5

Using cos^(x/2) = [1 + cos(x)] / 2

Then

2 * [1 + cos(x)] / 2 = cos(2x)  →

1 + cos(x) = cos(2x)  →

And using  cos(2x) = 2cos^2(x) - 1

1 + cosx = 2cos^2(x) - 1    →

2cos^2(x) - cosx - 2 = 0

let x = cosx

2x^2 -x - 2 = 0     .....using the on-site solver......

$${\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{2}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{17}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{4}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{17}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{4}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{0.780\: \!776\: \!406\: \!404\: \!415\: \!1}}\\
{\mathtt{x}} = {\mathtt{1.280\: \!776\: \!406\: \!404\: \!415\: \!1}}\\
\end{array} \right\}$$

So, using the cosine inverse, and ignoring the second answer, we have.....

cos-1 [(1 - √17)/4] = x ......  so x = 141.331°  and x = 218.668°

 

CPhill  Oct 22, 2014
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2+0 Answers

 #1
avatar+91475 
+5
Melody  Oct 22, 2014
 #2
avatar+81055 
+5
Best Answer

Using cos^(x/2) = [1 + cos(x)] / 2

Then

2 * [1 + cos(x)] / 2 = cos(2x)  →

1 + cos(x) = cos(2x)  →

And using  cos(2x) = 2cos^2(x) - 1

1 + cosx = 2cos^2(x) - 1    →

2cos^2(x) - cosx - 2 = 0

let x = cosx

2x^2 -x - 2 = 0     .....using the on-site solver......

$${\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{2}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{17}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{4}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{17}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{4}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{0.780\: \!776\: \!406\: \!404\: \!415\: \!1}}\\
{\mathtt{x}} = {\mathtt{1.280\: \!776\: \!406\: \!404\: \!415\: \!1}}\\
\end{array} \right\}$$

So, using the cosine inverse, and ignoring the second answer, we have.....

cos-1 [(1 - √17)/4] = x ......  so x = 141.331°  and x = 218.668°

 

CPhill  Oct 22, 2014

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