# Trig. equation

• Apr 27th 2014, 08:52 PM
jacks
Trig. equation
No. of solution of the equation $\displaystyle \cos \left(15\theta \right) = \cos \left(3\theta \right)$ ,where $\displaystyle \theta \in \left[0,2\pi\right]$
• Apr 27th 2014, 09:48 PM
Prove It
Re: Trig. equation
Using the identity \displaystyle \begin{align*} \cos{(X)} = \frac{e^{i\,X} + e^{-i\,X}}{2} \end{align*} we have

\displaystyle \begin{align*} \cos{(15x )} &= \cos{(3x )} \\ \frac{e^{15i\,x} + e^{-15i\,x}}{2} &= \frac{e^{3i\,x} + e^{-3i\,x}}{2} \\ e^{15i\,x} + e^{-15i\,x} &= e^{3i\,x} + e^{-3i\,x} \\ e^{3i\,x} \left( e^{15i\,x} + e^{-15i\,x} \right) &= e^{3i\,x} \left( e^{3i\,x} + e^{-3i\,x} \right) \\ e^{18i\,x} + e^{-12i\,x} &= e^{6i\,x} + 1 \\ e^{12i\,x} \left( e^{18i\,x} + e^{-12i\,x} \right) &= e^{12i\,x} \left( e^{6i\,x} + 1 \right) \\ e^{30i\,x} + 1 &= e^{18i\,x} + e^{12i\,x} \\ e^{30i\,x} - e^{18i\,x} - e^{12i\,x} + 1 &= 0 \\ \left( e^{6i\,x} \right) ^5 - \left( e^{6i\,x} \right) ^3 - \left( e^{6i\,x } \right) ^2 + 1 &= 0 \\ u^5 - u^3 - u^2 + 1 &= 0 \textrm{ if we let } u = e^{6i\,x} \end{align*}

\displaystyle \begin{align*} u = 1 \end{align*} is an obvious solution, so \displaystyle \begin{align*} (u - 1) \end{align*} is a factor. Long dividing gives

\displaystyle \begin{align*} \left( u - 1 \right) \left( u^4 + u^3 - u - 1 \right) &= 0 \end{align*}

\displaystyle \begin{align*} ( u - 1) \end{align*} is another obvious factor, so long dividing again gives

\displaystyle \begin{align*} \left( u - 1 \right) ^2 \left( u^3 + 2u^2 + 2u + 1 \right) &= 0 \end{align*}

\displaystyle \begin{align*} u = -1 \end{align*} is another solution, which means \displaystyle \begin{align*} (u + 1) \end{align*} is a factor. Long dividing again gives

\displaystyle \begin{align*} \left( u - 1 \right) ^2 \left( u + 1 \right) \left( u^2 + u + 1 \right) &= 0 \\ u = 1\textrm{ or } u = -1 \textrm{ or } u &= \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \\ u = 1 \textrm{ or } u = -1 \textrm{ or } u = -\frac{1}{2} - \frac{\sqrt{3}}{2}i \textrm{ or } u &= -\frac{1}{2} + \frac{\sqrt{3}}{2}i \\ u = e^{2k\, \pi \, i } \textrm{ or } u = e^{(2k + 1 ) \, \pi \, i } \textrm{ or } u = e^{\frac{2\pi}{3}k\, i } \textrm{ or } u &= e^{-\frac{\pi}{3}k\,i} \end{align*}

Note that k is an integer.

Case 1:

\displaystyle \begin{align*} u &= e^{2k\,\pi\,i} \\ e^{6i\,x} &= e^{2k\,\pi\,i} \\ 6i\,x &= 2k\,\pi \, i \\ x &= \frac{\pi}{3}k \\ x &= \left\{ 0, \frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3}, 2\pi \right\} \textrm{ if } x \in [0, 2\pi] \end{align*}

Case 2:

\displaystyle \begin{align*} u &= e^{(2k+1)\,\pi\,i} \\ e^{6i\,x} &= e^{(2k+1)\,\pi\,i} \\ 6i\,x &= (2k+1)\,\pi\,i \\ x &= \left( \frac{ \pi}{6} \right) \left( 2k + 1 \right) \\ x &= \left\{ \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{3\pi}{2} , \frac{11\pi}{6} \right\} \textrm{ if } x \in [0, 2\pi] \end{align*}

Case 3:

\displaystyle \begin{align*} u &= e^{\frac{2\pi}{3}\,k\,i} \\ e^{6i\,x} &= e^{\frac{2\pi}{3}\,k\,i} \\ 6i\,x &= \frac{2\pi}{3}\,k\,i \\ x &= \frac{\pi}{9}\,k \\ x &= \left\{ 0, \frac{\pi}{9}, \frac{2\pi}{9}, \frac{\pi}{3}, \frac{4\pi}{9} , \frac{5\pi}{9} , \frac{2\pi}{3}, \frac{7\pi}{9} , \frac{8\pi}{9} , \pi , \frac{10\pi}{9} , \frac{11\pi}{9} , \frac{4\pi}{3} , \frac{13\pi}{9} , \frac{14\pi}{9} , \frac{5\pi}{3}, \frac{16\pi}{9}, \frac{17\pi}{9} , 2\pi \right\} \textrm{ if } x \in [0, 2\pi]\end{align*}

Case 4:

\displaystyle \begin{align*} u &= e^{-\frac{\pi}{3}\,k\,i} \\ e^{6i\,x} &= e^{-\frac{\pi}{3}\,k\,i} \\ 6i\,x &= -\frac{\pi}{3}\,k\,i \\ x &= - \frac{\pi}{18}k \\ x &= \left\{ 0 , \frac{\pi}{18}, \frac{\pi}{9} , \frac{3\pi}{18}, \frac{2\pi}{9}, \frac{5\pi}{18}, \frac{\pi}{3}, \frac{7\pi}{18}, \frac{4\pi}{9}, \frac{\pi}{2}, \frac{5\pi}{9}, \frac{11\pi}{18}, \frac{2\pi}{3}, \frac{13\pi}{18}, \frac{7\pi}{9}, \frac{5\pi}{6}, \frac{8\pi}{9}, \frac{17\pi}{18} , \pi, \frac{19\pi}{18}, \frac{10\pi}{9}, \frac{7\pi}{6}, \frac{11\pi}{9}, \frac{23\pi}{18}, \frac{4\pi}{3}, \frac{25\pi}{18} , \frac{13\pi}{9} , \frac{3\pi}{2}, \frac{14\pi}{9} , \frac{29\pi}{18} , \frac{5\pi}{3}, \frac{31\pi}{18}, \frac{16\pi}{9}, \frac{11\pi}{6} , \frac{17\pi}{9}, \frac{35\pi}{18}, 2\pi \right\} \textrm{ if } x \in [0, 2\pi ] \end{align*}

I'll leave it to you to pull out each individual solution :)