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

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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.
  • #1
laprec
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Kindly assist with this question:
Determine the general solutions cos(2x+20)=-cos(x-11)
 
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  • #2
$\cos(2x+20)+\cos(x-11)=0$

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
 
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  • #3
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}
 
  • #4
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.
 

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  • #5
small correction ...
 

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  • #6
skeeter said:
small correction ...

Thank you very much! Much appreciated!
 

1. What is a trigonometric equation?

A trigonometric equation is an equation that involves one or more trigonometric functions (such as sine, cosine, or tangent) and an unknown variable. The goal is to solve for the value of the variable that makes the equation true.

2. How do I solve a trigonometric equation?

To solve a trigonometric equation, you can use algebraic techniques such as factoring, combining like terms, and applying trigonometric identities. It is also helpful to have a good understanding of the unit circle and the values of trigonometric functions for common angles.

3. What is the difference between solving for x and solving for theta in a trigonometric equation?

In a trigonometric equation, x and theta are both used to represent the unknown variable. The main difference is that x is typically used when dealing with angles measured in degrees, while theta is used when dealing with angles measured in radians.

4. Can you explain the steps for solving the equation cos(2x+20)=-cos(x-11)?

Sure! The first step is to use the double angle formula for cosine, which states that cos(2x) = 2cos^2(x) - 1. This will allow us to rewrite the equation as 2cos^2(x+10) - 1 = -cos(x-11). Then, we can use the Pythagorean identity for cosine, which states that cos^2(x) + sin^2(x) = 1. This will allow us to rewrite the equation as 2(1-sin^2(x+10)) - 1 = -cos(x-11). Simplifying this further, we get 2sin^2(x+10) - 1 = -cos(x-11). From here, we can use the identity cos(x) = sin(x+90) to rewrite the equation as 2sin^2(x+10) - 1 = sin(x+101). Finally, we can solve for sin(x) using the quadratic formula, and then use the inverse sine function to find the values of x that satisfy the equation.

5. Are there any special cases or restrictions when solving trigonometric equations?

Yes, there are a few special cases to keep in mind when solving trigonometric equations. These include the domain and range of the trigonometric functions, the periodic nature of trigonometric functions, and the fact that some equations may have multiple solutions within a given interval. It is important to carefully check your solutions and make sure they are within the appropriate range and satisfy the original equation.

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