+0  
 
0
159
3
avatar

Solve the following equation for x on the interval [0, 2pi]

2sin^3(4x)-sin(4x)=2cos^2(4x)-1

 Oct 22, 2018

Best Answer 

 #3
avatar+21867 
+10

Solve the following equation for x on the interval [0, 2pi]

2sin^3(4x)-sin(4x)=2cos^2(4x)-1

 

\(\begin{array}{|rcll|} \hline \mathbf{2\sin^3(4x)-\sin(4x)} &\mathbf{=}& \mathbf{2\cos^2(4x)-1} \\\\ \sin(4x)\Big(2\sin^2(4x)-1 \Big) &=& 2\cos^2(4x)-1 \\ -\sin(4x)\Big(1-2\sin^2(4x)\Big) &=& 2\cos^2(4x)-1 \\\\ && \boxed{\cos(8x)= 2\cos^2(4x)-1 } \\ && \boxed{\cos(8x)= 1-2\sin^2(4x) } \\\\ \mathbf{ -\sin(4x)\cos(8x) } & \mathbf{=} & \mathbf{\cos(8x)} \\ \hline \end{array}\)

 

Solution: \(\mathbf{ -\sin(4x)\cdot 0 = 0}\)

\(\begin{array}{|rcll|} \hline \mathbf{\cos(8x)} &\mathbf{=}& \mathbf{0} \\\\ 8x &=& \arccos(0)+\pi n,\qquad \mathbf{n \in Z} \\\\ 8x &=& -\dfrac{\pi}{2}+\pi n \quad & | \quad : 8 \\\\ x & = & \dfrac{\pi n}{8}-\dfrac{\pi}{16} \\\\ x & = & \pi\left( \dfrac{ n}{8}-\dfrac{1}{16} \right) \\\\ \mathbf{x} & \mathbf{=} & \mathbf{\pi\left( \dfrac{ 2n-1}{16}\right)} \\ \hline \end{array} \)

 

Solution: \(\mathbf{ -(-1)\cos(8x) = \cos(8x) }\)

\(\begin{array}{|rcll|} \hline \mathbf{\sin(4x)} &\mathbf{=}& \mathbf{-1} \\\\ 4x &=& \arcsin(-1)+2\pi n,\qquad \mathbf{n \in Z} \\\\ 4x &=& \dfrac{3\pi}{2} + 2\pi n \quad & | \quad : 4 \\\\ x &=& \dfrac{3\pi}{8} + \dfrac{\pi n}{2} \\\\ x & = & \dfrac{\pi n}{2}+\dfrac{3\pi}{8} \\\\ x & = & \pi\left( \dfrac{ n}{2}+\dfrac{3 }{8} \right) \\\\ \mathbf{x} & \mathbf{=} & \mathbf{\pi\left( \dfrac{ 4n+3}{8}\right)} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{\sin(4x)} &\mathbf{=}& \mathbf{\sin(\pi-4x) } = -1 \\\\ \pi-4x &=& \arcsin(-1)+2\pi n,\qquad \mathbf{n \in Z} \\\\ \pi-4x &=& \dfrac{3\pi}{2} + 2\pi n \\\\ 4x &=& 2\pi n-\dfrac{\pi}{2} \quad & | \quad : 4 \\\\ x & = & \dfrac{\pi n}{2}-\dfrac{ \pi}{8} \\\\ x & = & \pi\left( \dfrac{ n}{2}-\dfrac{ 1}{8} \right) \\\\ \mathbf{x} & \mathbf{=} & \mathbf{\pi\left( \dfrac{ 4n-1}{8}\right)} \\ \hline \end{array}\)

 

Solutions:

\(\begin{array}{|r|c|} \hline & x\text{ on the interval $[0, 2\pi]$} \\ \hline 1 & \dfrac{1}{16}\pi \\ \hline 2 & \dfrac{3}{16}\pi \\ \hline 3 & \dfrac{5}{16}\pi \\ \hline 4 & \dfrac{3}{8}\pi \\ \hline 5 & \dfrac{7}{16}\pi \\ \hline 6 & \dfrac{9}{16}\pi \\ \hline 7 & \dfrac{11}{16}\pi \\ \hline 8 & \dfrac{13}{16}\pi \\ \hline 9 & \dfrac{7}{8}\pi \\ \hline 10 & \dfrac{15}{16}\pi \\ \hline 11 & \dfrac{17}{16}\pi \\ \hline 12 & \dfrac{19}{16}\pi \\ \hline 13 & \dfrac{21}{16}\pi \\ \hline 14 & \dfrac{11}{8}\pi \\ \hline 15 & \dfrac{23}{16}\pi \\ \hline 16 & \dfrac{25}{16}\pi \\ \hline 17 & \dfrac{27}{16}\pi \\ \hline 18 & \dfrac{29}{16}\pi \\ \hline 19 & \dfrac{15}{8}\pi \\ \hline 20 & \dfrac{31}{16}\pi \\ \hline \end{array}\)

 

laugh

 Oct 23, 2018
edited by heureka  Oct 24, 2018
 #1
avatar
0

There should be 11 solutions

 Oct 22, 2018
 #2
avatar
0

Actually its 20 solutions.

Guest Oct 23, 2018
 #3
avatar+21867 
+10
Best Answer

Solve the following equation for x on the interval [0, 2pi]

2sin^3(4x)-sin(4x)=2cos^2(4x)-1

 

\(\begin{array}{|rcll|} \hline \mathbf{2\sin^3(4x)-\sin(4x)} &\mathbf{=}& \mathbf{2\cos^2(4x)-1} \\\\ \sin(4x)\Big(2\sin^2(4x)-1 \Big) &=& 2\cos^2(4x)-1 \\ -\sin(4x)\Big(1-2\sin^2(4x)\Big) &=& 2\cos^2(4x)-1 \\\\ && \boxed{\cos(8x)= 2\cos^2(4x)-1 } \\ && \boxed{\cos(8x)= 1-2\sin^2(4x) } \\\\ \mathbf{ -\sin(4x)\cos(8x) } & \mathbf{=} & \mathbf{\cos(8x)} \\ \hline \end{array}\)

 

Solution: \(\mathbf{ -\sin(4x)\cdot 0 = 0}\)

\(\begin{array}{|rcll|} \hline \mathbf{\cos(8x)} &\mathbf{=}& \mathbf{0} \\\\ 8x &=& \arccos(0)+\pi n,\qquad \mathbf{n \in Z} \\\\ 8x &=& -\dfrac{\pi}{2}+\pi n \quad & | \quad : 8 \\\\ x & = & \dfrac{\pi n}{8}-\dfrac{\pi}{16} \\\\ x & = & \pi\left( \dfrac{ n}{8}-\dfrac{1}{16} \right) \\\\ \mathbf{x} & \mathbf{=} & \mathbf{\pi\left( \dfrac{ 2n-1}{16}\right)} \\ \hline \end{array} \)

 

Solution: \(\mathbf{ -(-1)\cos(8x) = \cos(8x) }\)

\(\begin{array}{|rcll|} \hline \mathbf{\sin(4x)} &\mathbf{=}& \mathbf{-1} \\\\ 4x &=& \arcsin(-1)+2\pi n,\qquad \mathbf{n \in Z} \\\\ 4x &=& \dfrac{3\pi}{2} + 2\pi n \quad & | \quad : 4 \\\\ x &=& \dfrac{3\pi}{8} + \dfrac{\pi n}{2} \\\\ x & = & \dfrac{\pi n}{2}+\dfrac{3\pi}{8} \\\\ x & = & \pi\left( \dfrac{ n}{2}+\dfrac{3 }{8} \right) \\\\ \mathbf{x} & \mathbf{=} & \mathbf{\pi\left( \dfrac{ 4n+3}{8}\right)} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{\sin(4x)} &\mathbf{=}& \mathbf{\sin(\pi-4x) } = -1 \\\\ \pi-4x &=& \arcsin(-1)+2\pi n,\qquad \mathbf{n \in Z} \\\\ \pi-4x &=& \dfrac{3\pi}{2} + 2\pi n \\\\ 4x &=& 2\pi n-\dfrac{\pi}{2} \quad & | \quad : 4 \\\\ x & = & \dfrac{\pi n}{2}-\dfrac{ \pi}{8} \\\\ x & = & \pi\left( \dfrac{ n}{2}-\dfrac{ 1}{8} \right) \\\\ \mathbf{x} & \mathbf{=} & \mathbf{\pi\left( \dfrac{ 4n-1}{8}\right)} \\ \hline \end{array}\)

 

Solutions:

\(\begin{array}{|r|c|} \hline & x\text{ on the interval $[0, 2\pi]$} \\ \hline 1 & \dfrac{1}{16}\pi \\ \hline 2 & \dfrac{3}{16}\pi \\ \hline 3 & \dfrac{5}{16}\pi \\ \hline 4 & \dfrac{3}{8}\pi \\ \hline 5 & \dfrac{7}{16}\pi \\ \hline 6 & \dfrac{9}{16}\pi \\ \hline 7 & \dfrac{11}{16}\pi \\ \hline 8 & \dfrac{13}{16}\pi \\ \hline 9 & \dfrac{7}{8}\pi \\ \hline 10 & \dfrac{15}{16}\pi \\ \hline 11 & \dfrac{17}{16}\pi \\ \hline 12 & \dfrac{19}{16}\pi \\ \hline 13 & \dfrac{21}{16}\pi \\ \hline 14 & \dfrac{11}{8}\pi \\ \hline 15 & \dfrac{23}{16}\pi \\ \hline 16 & \dfrac{25}{16}\pi \\ \hline 17 & \dfrac{27}{16}\pi \\ \hline 18 & \dfrac{29}{16}\pi \\ \hline 19 & \dfrac{15}{8}\pi \\ \hline 20 & \dfrac{31}{16}\pi \\ \hline \end{array}\)

 

laugh

heureka Oct 23, 2018
edited by heureka  Oct 24, 2018

6 Online Users

avatar