# Solve trig equation cos(2x+20)=-cos(x-11)

• MHB
• laprec
In summary, the conversation discusses finding the general solutions for the equation $\cos(2x+20)+\cos(x-11)=0$ using the sum to product identity. The conversation also presents an alternative method using the unit circle to determine the solutions, which involves considering the arguments of the cosine functions and adding multiples of $360^\circ$ to account for the period of the circle. The final output also includes a small correction made by one of the speakers.

#### laprec

Kindly assist with this question:
Determine the general solutions cos(2x+20)=-cos(x-11)

Last edited:
$\cos(2x+20)+\cos(x-11)=0$

using sum to product identity

$2\cos\left(\dfrac{3x+9}{2}\right)\cos\left(\dfrac{x+31}{2}\right) =0$

setting each cosine factor equal to zero yields

$\dfrac{3x+9}{2} = \dfrac{\pi}{2}(2k+1)$ where $k \in \mathbb{Z}$

$\dfrac{x+31}{2} = \dfrac{\pi}{2}(2k+1)$ where $k \in \mathbb{Z}$

if the arguments of the cosine functions are in degrees, then

$\dfrac{3x+9}{2} = 90(2k+1)$ where $k \in \mathbb{Z}$

$\dfrac{x+31}{2} = 90(2k+1)$ where $k \in \mathbb{Z}$

finish it

Last edited by a moderator:
As an alternative to skeeter's method, I'd like to bring up the unit circle.
\begin{tikzpicture}[scale=3,>=stealth]
\def\angle{35}
\draw[->, help lines] (-1.2,0) -- (1.2,0);
\draw[->, help lines] (0,-1.2) -- (0,1.2);
\draw[ultra thick, blue] circle (1);
\draw[thick] (0,0) -- (\angle:1) -- ({cos(\angle)}, 0) node[below] {$\cos\theta$} -- cycle ;
\draw[thick] (0,0) -- ({180+\angle}:1) -- ({-cos(\angle)}, 0) node[below] {$-\cos\theta$} -- cycle;
\draw[thick] (0,0) -- ({180-\angle}:1) -- ({-cos(\angle)}, 0);
\draw[->] ({\angle/2}:.4) node {$\theta$} (0:.3) arc (0:\angle:.3);
\draw[->] ({(180+\angle)/2}:-.14) node {$180^\circ+\theta$} (0:.1) arc (0:{180+\angle}:.1) ;
\draw[->] ({(180-\angle)/2}:.29) node {$180^\circ-\theta$} (0:.2) arc (0:{180-\angle}:.2) ;
\end{tikzpicture}
We have an equation of the form $\cos\theta = -\cos\phi$.
Given a $\theta$, for $\cos\theta$ to be equal to the opposite of another cosine, we can see that the other angle must either be $180^\circ-\theta$ or $180^\circ+\theta$. And we may have to add a multiple of $360^\circ$, which is the period of the circle.

So:
\begin{array}{lcl}
\cos(2x+20^\circ)=-\cos(x-11^\circ) \\
2x + 20^\circ = 180^\circ - (x-11^\circ) + 360^\circ k &\lor& 2x + 20^\circ = 180^\circ + (x-11^\circ) + 360^\circ k \\
3x = 171^\circ + 360^\circ k &\lor& x = 149^\circ + 360^\circ k \\
x = \frac 13\cdot171^\circ + 120^\circ k &\lor& x = 149^\circ + 360^\circ k \\
\end{array}

skeeter said:
$\cos(2x+20)+\cos(x-11)=0$

using sum to product identity

$2\cos\left(\dfrac{3x+9}{2}\right)\cos\left(\dfrac{x+31}{2}\right) =0$

setting each cosine factor equal to zero yields

$\dfrac{3x+9}{2} = \dfrac{\pi}{2}(2k+1)$ where $k \in \mathbb{Z}$

$\dfrac{x+31}{2} = \dfrac{\pi}{2}(2k+1)$ where $k \in \mathbb{Z}$

if the arguments of the cosine functions are in degrees, then

$\dfrac{3x+9}{2} = 90(2k+1)$ where $k \in \mathbb{Z}$

$\dfrac{x+31}{2} = 90(2k+1)$ where $k \in \mathbb{Z}$

finish it
Thanks a million skeeter. Much appreciated!
I have attached complete work out based on your guidiance.

- - - Updated - - -

Klaas van Aarsen said:
As an alternative to skeeter's method, I'd like to bring up the unit circle.
\begin{tikzpicture}[scale=3,>=stealth]
\def\angle{35}
\draw[->, help lines] (-1.2,0) -- (1.2,0);
\draw[->, help lines] (0,-1.2) -- (0,1.2);
\draw[ultra thick, blue] circle (1);
\draw[thick] (0,0) -- (\angle:1) -- ({cos(\angle)}, 0) node[below] {$\cos\theta$} -- cycle ;
\draw[thick] (0,0) -- ({180+\angle}:1) -- ({-cos(\angle)}, 0) node[below] {$-\cos\theta$} -- cycle;
\draw[thick] (0,0) -- ({180-\angle}:1) -- ({-cos(\angle)}, 0);
\draw[->] ({\angle/2}:.4) node {$\theta$} (0:.3) arc (0:\angle:.3);
\draw[->] ({(180+\angle)/2}:-.14) node {$180^\circ+\theta$} (0:.1) arc (0:{180+\angle}:.1) ;
\draw[->] ({(180-\angle)/2}:.29) node {$180^\circ-\theta$} (0:.2) arc (0:{180-\angle}:.2) ;
\end{tikzpicture}
We have an equation of the form $\cos\theta = -\cos\phi$.
Given a $\theta$, for $\cos\theta$ to be equal to the opposite of another cosine, we can see that the other angle must either be $180^\circ-\theta$ or $180^\circ+\theta$. And we may have to add a multiple of $360^\circ$, which is the period of the circle.

So:
\begin{array}{lcl}
\cos(2x+20^\circ)=-\cos(x-11^\circ) \\
2x + 20^\circ = 180^\circ - (x-11^\circ) + 360^\circ k &\lor& 2x + 20^\circ = 180^\circ + (x-11^\circ) + 360^\circ k \\
3x = 171^\circ + 360^\circ k &\lor& x = 149^\circ + 360^\circ k \\
x = \frac 13\cdot171^\circ + 120^\circ k &\lor& x = 149^\circ + 360^\circ k \\
\end{array}

Thanks a lot Klaas van Aarsen, the alternative method is equally helpful and insightful. Much appreciated.

#### Attachments

• mhboardsoln.PNG
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small correction ...

#### Attachments

• cos_equation.jpg
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skeeter said:
small correction ...

Thank you very much! Much appreciated!